**Q.1. What is Boyle’s law of gases?**

Ans. This law states, the volume of a given mass of a gas is inversely proportional to the pressure at constant temperature. Mathematics Boyle’s law can be expressed as:

V (When T and n are constant)

Or

V = Or PV = K (When T and n are constant)

Where ‘k’ is proportionality constant. The value of k is different for the different amounts of the same gas.

**Q.2. What are isotherms? What happens to the position s of isotherms when they are plotted at high temperature for a particular gas?**

Ans. The curves obtained, when a graph is plotted between pressure and volume at constant temperature are called isotherms, ‘iso’ means same ‘therm’ means heat.

Isotherms go away from both the axes when they are plotted at high temperature for a particular gas, since the volume of the gas has increased at higher temperature.

**Q.3. Why do we get a straight line when pressure exerted on a gas are plotted against inverse of volumes? This straight line changes its position in the graph by varying the temperature.**

Ans. When the pressure exerted on a gas is plotted against I/V, we get a straight line because pressure and inverse of volume (I/V) are directly proportional to each other at constant temperature. This straight line changes its position by varying the temperature, because this straight line at higher temperature will be close to zero; which means when P is very close to zero, then the volume is so high that I/V is very close to zero;.

**Q.4. The product of pressure and volume of a gas at constant temperature and number of moles is a constant quantity . why? **

Ans. According to Boyle’s law, when the temperature and number of moles of a gas are constant, then the increase in pressure will decrease the volume in proportion to the increase, so that the product of pressure and volume remains constant (PV = K). by doubling the pressure the volume becomes half. Thus

P_{1}V_{1} = P_{2}V_{2} = P_{3}V_{3} = K

**Q.5. How will you explain that the value of the constant k in the equation PV depends upon (i) the temperature (ii) the quantity of a gas.**

Ans. In Boyle’s law, the temperature and number of moles (quantity) of an ideal gas are kept constant, so that the value of the product PV remains constant. Now, if the temperature of the gas is increased, then the volume increase for the same quantity of the gas and thus the value of product PV increases.

When the quantity of gas is increased at the same temperature, then the volume of gas increases and thus the value of product PV increases. From these observations, it is clear that the value of the constant k in equation PV depends upon the temperature and the quantity of a gas.

**Q.6. What is the Charles’s law? Which scale of temperature is used to verify V/T = K (P and n are constant)?**

Ans. Charles’s law gives the relationship between gas volume and temperature. This law states, the volume of a given mass of a gas is directly proportional to the absolute temperature when the pressure is kept constant. Mathematically

V ∝ T Or V = kT (When P and n are constant)

Or = K (When P and n are constant)

Thus, doubling the absolute temperature, causes the gas volume of double. If the temperature is changed from T_{1} to T_{2}, the volume changes from V_{1} to V_{2}, then

and or

The value of the constant, k, depends on the pressure and amount of the gas. Kelvin temperature scale is used to verify that V/T = K.

**Q.7. What is absolute zero?**

Ans. the hypothetical temperature, -273.16^{o}C at which the volume of the gas would become zero is called absolute zero, and is taken as the zero point on the kelvin scale temperature. Of course, this temperature, -273.16 is never achieved because all gases liquefy or solidify before reaching this temperature. Thus real gases will liquefy or solidify while approaching absolute zero. For routine calculations, the value of absolute zero is taken as -273^{o}C.

**Q.8. A gas occupies 100dm ^{3} at 283K. what will be its volume at -273^{o}C at constant pressure? **

Ans. All real gases liquefy or solidify before reaching this temperature, -273^{o}C. it means that the gas is liquefied or solidified before reaching -273^{o}C, so the volume of the gas cannot be predicted at -273^{o}C at constant pressure.

**Q.9. Justify that the volume of a given mass of a gas becomes theoretically zero at -273 ^{o}C. or give the quantitative definition of Charles law.**

Ans. The quantitative definition of Charles’s law is “ At constant temperature, the volume of a given mass of gas increase or decreases by 1/273 of its original volume at 0^{o}C for every 1^{o}C rise or fall In temperature” according to this definition, if the volume of the gas at 0^{o}C is 273 cm^{3}. Now if the temperature is decreased by 1^{o}C, the volume of the gas decreases by x 273 cm^{3} or 1 cm^{3}; the final volume is thus 273 cm^{3} – 1 cm^{3} = 272 cm^{3}. At -273^{o}C, the volume of the gas would become theoretically zero. The general equation to know the volumes of the gas at various temperature is

V_{t} = V_{o} (1 + )

When T = -273^{o}C; then V _{-273}^{o}C = Vo (1 – ) = 0

Thus the volume of a give mass of a gas becomes theoretically zero at -273^{o}C.

**Q.10. What is Kelvin scale of temperature?**

Ans. A scale of temperature at which -273.15^{o}C, is taken as zero Kelvin is called Kelvin scale temperature. Zero on the Kelvin scale corresponds to -273^{o}C. therefore K = 0^{o}C + 273.15. the Kelvin scale, however, is the SI temperature scale and the SI unit of temperature is the Kelvin (K).

**Q.11. Throw some light on the factor 1/273 in Charles’s law?**

Ans. The factor 1/273 in Charles’s law shows that at constant pressure the volume of a given mass of a gas increases or decrease by 1/273 of its original volume at 0^{o}C for every rise or fall in temperature respectively. The general equation to know the volumes of the gas at various temperatures is

V_{t} = V_{o} (1 + )

Where Vt is the volume of the gas at t^{o}C, V_{o} is the volume of the gas at 0^{o}C and t is the temperature on the centigrade scales.

**Q.12. Do you think that the volume of any quantity of a gas becomes zero at -273.16 ^{o}C. is not against the law of conservation of mass? How do you deduce the idea of absolute zero from this information?**

Ans. The volume occupied by a gas does not become zero at -273^{o}C, because all the gases will liquefy before reaching this temperature -273^{o}C. it is against the law of conservation of mass that the volume occupied by the gas becomes zero. This hypothetical temperature -273^{o}C, at which the volume of a given mass of a gas becomes zero is called **absolute zero.**

**Q.13. What is the general gas equation? Derive it?**

Ans the equation which gives simultaneously the effect of changes of pressure and temperature on the volume of a given mass of a gas is known as the general gas equation or ideal gas equation. The general gas equation is derived by combining Boyle’s law, Charles’s law and Avogadro’s law.

According to Boyle’s law: V (when T and n are constant)

According to Charles’s law V T (when P and n are constant)

According to Avogadro’s law V n (at constant T and P)

If none of the variables are kept constant, then combining the above three relations, we get.

V Or V = R or PV = nRT

Where R is called the general gas constant and the equation, PV = nRT, is called general gas equation.

**Q.14. How the value of the general gas constant ‘R’ can be derived with the help of Avogadro’s law?**

Ans. According to Avogadro’s law, the volume of one mole of an ideal gas at STP (one atmospheric pressure and 273.16 K) is 22.414 dm^{3}. Putting these values of P, T, V and n, in the general gas equation will give the value of R.

R = = = 0.0821 dm^{3} atm K^{-1} mol ^{-1}

**Q.15. Calculate the value of R in SI units. **

Ans. Using SI units of pressure and volume in the general gas equation, the value of R can Be calculated as follows. The SI units of pressure are Nm^{-2} and of volume is m^{3}.

1 atm = 760 torr = 101325 Nm^{-2}; V = 0.022414dm^{3} (1m^{3} = 1000 dm^{3})

T = 273.15 K; n = 1 mol

R = = = 8.3143 NmK^{-1} mole^{-1} = 8.3143 JK^{-1} mol^{-1}

Note: 1Nm = 1J ; the units of R are expressed in energy K^{-1} mol^{-1}. 1 erg = 10^{-7} Joule ; so R = 83.143 x 10^{-7} erg K^{-1} mol^{-1}

** Q.16. How will you calculate the density of an ideal gas from the general gas equation?**

Ans. The density of an ideal gas can be calculated from the general gas equation by substitution the value of number of moles, n, terms of the mass, m, and the molar mass, M, of the gas.

PV = nRT = RT (n = )

Or PM = RT

PM = dRT (d = )

Or d =

**Q.17. How the density of an ideal gas double by doubling the pressure or decreasing the temperature on Kevin scale by ½ ?**

Ans. We Know that d=PM/RT

The density of an ideal gas is directly proportional to the pressure on the gas and is inversely proportional to the absolute temperature. Thus by doubling the pressure on the gas, the density becomes double. Similarly, when the temperature becomes one half, the density becomes double.

**Q.18. How do you justify from general equation that increase in temperature or decrease of pressure decrease the density of the gas?**

Ans. The equation for the density of a gas is d = PM/RT< which is derived from the general gas equation, PV = nRT . according to this equation, density is directly proportional to the pressure and is inversely proportional to the absolute temperature, therefore, increases of temperature of decrease of pressure decreases the density of the gas.

**Q.19. What is Avogadro’s law of gases?**

Ans. This law states, “equal volumes of all the ideal gases at the same temperature and pressure contain equal number of molecules”. Since equal number of moles of different ideal gases at the same temperature and pressure contain equal number of molecules, therefore, the number of moles on of any ideal gas is directly proportional to its volume, V. So

V ∝ n

One mole of an ideal gas at 273.16K and 1 atm pressure has a volume of 22.414dm^{3}. One mole of a gas has Avogadro’s number of particles, so 22.414dm^{3} of various ideal gases will have Avogadro’s number of molecules, i.e. 6.02 x 10^{23}.

**Q.20. Do you think that 1 mole of H _{2} and 1 mole of NH_{3} at 0^{o}C and one atm pressure will have Avogadro’s number of particles.**

Ans. One mole of H_{2} of 0^{o}C and atm pressure will have Avogadro’s number of particles, since H_{2} gas at 0^{o}C and 1 atm pressure behaves like an ideal gas. Whereas, 1 mole of NH_{3} at 0oC and 1 atm pressure will not have Avogadro’s number of particles, because it does not behave like na ideal gas due to the presence of dipole interactions and hydrogen bonding among its molecules.

**Q.21. Justify that 1cm _{3} of H_{2} an d1 cm^{3} of CH_{4} at STP will have same number of molecules. Although one molecule of CH_{4} is 8 times heavier than that of hydrogen.**

Ans. According to Avogadro’s law, “Equal volumes of ideal gases at the same temperature and pressure contain equal number of molecules. Both H_{2} and CH_{4} is STP behave like ideal gases and both have equal volumes (1 Cm^{3} each), therefore, 1 cm^{3} of H_{2} and 1 cm^{3} of CH_{4} at STP will have same number of molecules. Although, one molecule of CH_{4} is 8 times heavier than that of H_{2} but this does not disturb the volume occupied, because molecules of the gases are widely separated from each other at because molecules of the gases are widely separated from each other at STP.

**Q.22. State Dalton’s law of partial pressures. Gives its expression. **

Ans. This law states, the total pressure exerted by a mixture of non-reacting gases is equal to the sum of their individual partial pressure. Let P_{t} be the total pressure and P_{1}, P_{2}, P_{3}, ……….. be the partial pressures of the gases in the mixture. Then

P_{t} = P_{1}+ p_{2} + p_{3}

Since each gas in the mixture behaves independently. So the general gas equation (PV = nRT) is applicable to the individual gases in the gaseous mixture. Let n_{1}, n_{2}, and n_{3} be the number of moles of each of the gas in the mixture and n_{t} be the total number of gaseous mixture

(n_{t} = n_{1} +n_{2}+n_{3}). Then

P_{1} = n_{1} ; p2 n_{2} ; p_{3} = n_{3}

P_{t} = (n_{1} +n_{2}+n_{3}) = n_{t}

We can relate the amount of a give gas in a mixture to its partial pressure by dividing the above equations.

The ratio n_{1}/n_{t} is called the mole ratio of gas 1, denoted X_{t}, then

or

**Q.23. What is meant by partial pressure of gas?**

Ans. The partial pressure of a gas in a mixture of gases is the pressure that it would exert on the walls of the container, if it were present all alone in the same volume under the same temperature.

**Q.24. Explain that the process of respiration obeys the Dalton’s law of partial pressure.**

Ans. The process of respiration depends upon the difference in partial pressures and is the application of Dalton’s law partial pressure. When human beings inhale air then oxygen moves into lungs as the partial pressure of oxygen in air (159 torr) is more than that of the partial pressure of oxygen in lungs (116 torr). CO_{2} produced during respiration moves out from the lungs into air, as the partial pressure of CO_{2} is more in the lungs than that in air. Thus, process of respiration obeys the Dalton’s law of partial pressure.

**Q.25. Dalton’s law of partial pressures is only obeyed by those gases which do not have attractive forces among their molecules (ideal gases), explain it.**

Ans. Gas pressure on a wall is due to molecules colliding with it . when a molecule about to collide with the wall, it is attracted away from the wall by attractive force of neighboring molecules (intermolecular forces). As a result, the pressure exerted by the gas would be less than that of an ideal gas. Therefore, gases having attractive forces among their molecules do not obey Dalton’s law of partial pressures, which is only obeyed by ideal gases.

**Q.26. At higher altitudes, the pilots feel uncomfortable breathing. Why?**

Ans. At higher altitudes, the pilots feel uncomfortable breathing because the partial pressure of oxygen in the unpressurizatoin cabin is low as compared to 159 torr, where one feels comfortable breathing.

**Q.27. how will you calculate the partial pressure of dry gas which is collected over water?**

Ans. When a gas is collected over water, it becomes moist. The pressure exerted by this moist gas is, therefore, sum of the partial pressures of the dry gas and that of water vapour. Mathematically,

P_{moist} = d_{dry} + P water vapour

P_{moist} = d_{dry} + aqueous tension

P_{dry} = P_{moist} – aqueous tension

The partial pressure exerted by the water vapours is called aqueous tension.

**Q.28. Differentiate between diffusion and effusion of gases?**

Ans. The spontaneous mixing of the molecules of different gases by random motion and collisions to form homogeneous mixture is called gaseous diffusion. Th escape of gas molecules one by one without collisions through a pin hole in their container into an evacuated space is called effusion. The diffusion is the spread of one substance throughout a space or throughout second substances.

**Q.29. What is Graham’s law of diffusion?**

Ans. This law states that the rate of diffusion or effusion of a gas is inversely proportional to the spare root of its density at constant temperature and pressure.

Rate of diffusion (at constant T and P)

Rate of diffusion = or rate x = K

Let two gases 1 and 2, having rates of diffusion as r_{1} and r_{2} and densities as d_{1} and d_{2} respectively. Then according to Graham’s law of diffusion

r_{1} x = K

r2 x = K

Dividing the above two equations and rearranging

Since density of a given gas is directly proportional to its molecular mass. So, Graham’s law of diffusion can be written as

Where M_{1} and M_{2} are the molecular masses of gases:

**Q.30. Lighter gases diffuse rapidly than heavier gases. Give reason. **

Ans. Although the average kinetic energies of different gases are the same at the same temperature, but their molecular masses are different, so their velocities will also be different at the same temperature. The lighter gas molecules would have greater velocities and so they diffuse rapidly than heavier gas molecules.

**Q.31. The rate of diffusion of NH _{3} is 1.5 times greater than HCl. Explain.**

Ans. According to Graham’s law of diffusion, the rates of diffusion of gases are inversely proportional to the square roots of their molecular masses. The molecular mass of HCl is 36.5 amu and that of NH_{3} is 17 amu. Thus

This shows that the rate of diffusion of NH_{3} Is 1.5 times greater than HCl.

**Q.32. What is kinetic equation for ideal gases?**

Ans. R.J. Clausius deduced an expression for the pressure of an ideal gas from the basic assumptions of kinetic theory of gases in the form of an equation known as kinetic equation. According to kinetic molecular theory, the pressure of a gas is the result of collisions of molecules with the walls of the container. Due to these collisions, a force is exerted on the walls of the container. The force when divided by the area of the vessel gives force per unit area, which is called pressure. The kinetic equation is

PV =

Where P = pressure, V = volume, m = mass of one molecule of the gas, N = number of molecules of gas in the vessel and = mean square

**Q.33. What is man square velocity .**

Ans. It is the average of the squares of all the possible velocities of gas molecules. If there are n_{1} molecules with velocity c_{1}, n_{2} molecules with velocity c_{2}, and so on, then,

=

Here, n_{1} +n_{2} +n_{3} + ——— = N

**Q.34. What is root mean square velocity, c _{rms.}**

Ans. The root mean square velocity is defined as the under-root of the mean of the squares of the different velocities of all the N molecules of the gas.

C_{rms} =

The expression for the root mean square velocity deduced from the kinetic equation is

C_{rms} = (M = molar mass of the gas)

**Q.35. How will you derive Boyle’s law from kinetic gas equation?**

Ans. According to the kinetic theory of gases, the kinetic energy of gas molecules is directly proportional to the absolute temperature of gas. Thus

T Or = kT (k is the proportionality constant)

According to the kinetic gas equation, PV

Multiplying the kinetic gas equation by , we get PV = ( ) =

But = kT

Hence PV = kT Or PV = kT

Since k is constant, so at constant temperature, PV is Constant, which is Boyle’s law.

**Q.36. How will you derive Charles’s law from kinetic gas equation?**

Ans. According to kinetic theory of gases, the kinetic energy of gas molecules, , is directly proportional to the absolute temperature of the gas. Thus

T Or = kT

We know. PV = , (kinetic equation)

Multiplying the kinetic equation by , we get ( ) = PV = kT (since = kT)

Or PV = kT Or V = T

Since k is constant, so at constant pressure, V T which is Charles’s law.

**Q.37. Derive Avogadro’s law from kinetic theory of gas equation?**

Ans. Consider equal volumes of two gases 1 and 2 under the same conditions of temperature and pressure, having molecules N_{1} and N_{2}, masses m_{1} and m_{2} and mean square velocities respectively. Then the kinetic equation for two gases can be written as

PV = (Kinetic equation for gas 1)

PV = (Kinetic equation for gas 2)

Equalizing= PV = = PV = Or (1)

At the same temperature, the mean kinetic energies per molecule of both gases will be the same, so

= PV = Or (2)

Dividing equation (1) by Eq. (2) we get

N_{1} = N_{2}

Thus equal volumes of all the gases at the same temperature and pressure contain equal number of molecules, which is Avogadro’s law.

**Q.38. Derive Graham’s law of diffusion from kinetic theory of gases?**

Ans. Applying the kinetic equation PV =

For 1 mole of an ideal gas, N = N_{A} (Where NA is Avogadro’s number),

Hence PV = Or PV = (M = mN_{A} = molecular mass of gas)

Or (Since = d, density of gas)

Or

Since the root mean square velocity of the gas is directly proportional to the rate of diffusion of the gas. Thus r

So r

At constant pressure, r , which is Graham’s law.

**Q.39. Describe briefly the kinetic interpretation f temperature?**

Ans. According to one fundamental postulate of kinetic theory, the absolute temperature of a gas is a measure of the average kinetic energy of its molecules. However, according to equation, E_{k} = 3RT/2N_{A}, the kelvin (absolute) temperature of a gas is directly proportional to the average translational kinetic energy of its molecules. It means a change in temperature means change in the intensity of molecular motion. Thus at absolute zero of temperature the molecular motions cease. The absolute zero is unattainable, however, current attempts have resulted in temperature as low as 10^{-5}K. in gases and liquids, absolute temperature is the measure of average translational kinetic energies of molecules, while in solids, it becomes a measure of vibrational kinetic energy.

**Q.40. What two properties of gas molecules cause them to behave non-ideally?**

Ans. The real (actual) volume of gas molecules and attractive forces between molecules cause gases to behave non-ideally.

**Q.41. Under what conditions of temperature and pressures do gases usually behave non-ideally?**

Ans. Non-ideal gas behaviors is observed at very high pressure and low temperatures.

**Q.42. Give two causes for deviation of gases from ideality. Or what are faulty points in kinetic molecular theory of gases?**

Ans. The deviation of real gases from ideality is due to the following two faulty assumptions.

- The actual volume of gas molecules is negligible as compared to the volume of the vessel.
- There are no forces of attraction among the molecules of a gas.

Real gas molecules, however, do have finite volumes, and they do attract one another.

**Q.43. Why do real gases deviate from the ideal behavior at low temperature and pressure?**

Ans. As the temperature of a gas is lowered, the average kinetic energy decreases and thus the attractive forces between the molecules become significant (dominant), and thus real gases deviate from the ideality at low temperature.

At high pressure, the gas molecules are pushed closer together and the volume of molecules themselves become significant relative to total volume occupied by the gas. So, collisions between molecules and attractive forces become significant. Thus at high pressure real gases deviate from ideal behaviors.

**Q.44. Why gases do not settle?**

Ans. According to the kinetic molecular theory of gases, the gas molecules are in a constant motion, they collide with one another and their collisions are perfectly elastic i.e. there is no gain or loss in energy during their collisions. Therefore, gases do not settle.

**Q.45. Briefly describe the significance of the constants ‘a’ and ‘b’ in the van der Walls equation.**

Ans. The van der walls constant ‘a’ is a measure of the intermolecular attractive forces. The van der Waals constant ‘b’ is a measure of actual volume occupied by a mole of gas molecules. The values of both ‘a’ and ‘b’ constants generally increase with an increase in mass of molecule and with an increase in the complexity of its structure. The units of ‘a’ is pressure (volume)^{2} mol^{-2}, i.e., atm dm^{6} mol^{-2}. Is SI units ‘a’ is expressed Nm^{-2} x (m^{3})^{2} mol^{-2} or Nm^{4}mol^{-2}, while b is a volume mol^{-1}, i.e., dm^{3} mol^{-1}.

**Q.46. What is van der Waals equation of state?**

Ans. Van der Waals recognized that for a real gas, the expression for an ideal gas, PV = nRT/V would have to be corrected for the two effects: the finite volume occupied by the gas molecules, and the attractive forces between the gas molecules. He introduced two constants ‘a’ and ‘b’ to make these corrections.

The volume is decreased by the factor nb, which accounts of the finite volume occupied by the gas molecules. The pressure is in turn decreased by the factor na/V^{2}, which accounts for the attractive forces between the gas molecules. Correcting the pressure and volume terms in an ideal gas equation, we obtain van der Waals equation.

(P +n^{2}a/V^{2} ) (V – nb) = nRT (Van der Waals equation)

**Q.47. Pressure of NH _{3} gas at given conditions (say 20 atm pressure and room temperature) is less as calculated by van der Waals equation than that calculated by general gas equation. Explain.**

Ans. NH_{3} is a polar molecule and attractive forces exist among its molecules. These attractive forces lessen the force with which the molecule hits the wall. As a result, the pressure is less than that of an ideal gas. Moreover. NH_{3} is a real gas, do have finite volume and , therefore, the less than the observed volume. Clearly, the ideal volume is less than the observed volume. Van der Waal’s introduced these two corrections in the ideal gas equation. Therefore, pressure of NH_{3} gas is less as calculated by van der Waal’s equation than that calculated by general gas equation.

**Q.48. Water vapours do not behave ideally at 273K. explain.**

Ans. Water vapours at 273^{o}C have sufficient inter-molecular forces, because 273^{o}C is below the critical temperature of H_{2}O. therefore, water vapours do not behave ideally.

**Q.49. SO _{2} is comparatively non-ideal at 273 K but behaves ideally at 327^{o}C. Explain.**

Ans. SO_{2} is non-ideal at 273K or 0^{o}C, because this temperature is low and inter-molecular forces exist among SO_{2} molecules. While 327^{o}C is very high temperature for SO_{2} gas and there are negligible intermolecular forces among gas molecules. Therefore, SO_{2} behaves ideally at 327^{o}C.

**Q.50. Gases deviate more from the ideal behavior at 0 ^{o}C then at 100^{o}C.**

Ans. At 0oC (low temperature) the attractive forces between the gas molecules are significant (dominant) and thus gases become non-ideal. At 100^{o}C (high temperature), the forces of attractions are negligible and thus they behave ideally.

**Q.51. Hydrogen and He are ideal at room temperature but SO _{2} and Cl_{2} are non-ideal. How do you explain it?**

Ans. H_{2} and He have very low boiling points which are far below the room temperature. So, the room temperature is very high for H_{2} and He. Moreover, the molecules of H_{2} and He have very small masses. Due to high temperature and small masses of H_{2} and He, the intermolecular attractions become negligible and they behave ideally while the boiling points of SO_{2} and Cl_{2} are close to room temperature and their molecular masses are relatively high. Thus sufficient attractive forces are present among their molecules; therefore, SO_{2} and Cl_{2} are non-ideal gases.

**Q.52. what is Joule-Thomson effect? Joule Thomson effect is operative in the Linde’s method of liquefaction of air. How?**

Ans. When a compressed gas is allowed to expand into a region of low temperature, it produces intense cooling and this phenomenon is called Joule-Thomson effect. In Linde’s method, the gas is compressed to about 200 atmosphere and this compressed gas is suddenly allowed to expand through and nozzle. The free expansion of the gas results in a fall in temperature. The process is repeated again and against until the gas is liquefied. So, in this way, Joule Thomson effect is operative in the Linde’s method of liquefaction of air.

**Q.53. What is critical temperature of a gas? What is its importance for liquefaction of gases?**

Ans. The temperature above which a gas cannot be liquefied, no matter how much pressure is applied is called critical temperature. In other words, the highest temperature at which a substance can exist as a liquid, is called its critical temperature. The critical temperature and the critical pressure provides us the information about the condition under which gases liquefy.

**Q.54. What is Plasma state?**

Ans. Plasma is often called the “fourth state of matter”. It occurs only in lightening discharges and in artificial devices like fluorescent lights, neon signs, etc. the lionized gas mixture, consisting of ions, electrons and neutral atoms is called plasma. Plasma is a distinct state of matter containing a sufficient number of electrical charged particles to affect its electrical and magnetic properties and behaviors.

**Q.55. H _{2} diffuses through a porous plate at a rate of 500cm^{3} per minute at 0^{o}C. What is the rate of diffusion of O_{2} through the same porous plate at 0^{o}C?**

Ans. Molar mass of H_{2} = 2 g mol^{-1} ; molar mass of O_{2} = 32 g mol^{-1}

Min^{-1}

**Q.56. why some amount of pressure should be added in the measured pressure of the non-ideal gas to get the ideal pressure of the gas?**

Ans. In an ideal gas, there are no attractive forces between molecules and the molecular bombardment on the walls of vessel results in the ideal pressure, P_{ideal}. However, in non-ideal gases there are forces of attraction between the molecules and become significant at higher pressures, the molecules will not strike the walls quite as hard on the average because each molecule will be slightly held back by the other nearby molecules. Thus the measured pressure P is slightly less than P_{ideal}. So van der Waal needed a pressure correction, therefore lessened pressure is added in the measured pressure to get the ideal pressure : p_{i} = P + P’

**Q.57. What are the characteristics of plasma?**

Ans. 1. The motions of the particles in the plasma generate fields and electric current from within plasma density of charged particles. This complex set of interactions makes plasma a unique, fascinating and complex state of matter.

- Although plasma includes electrons and ions and conducts electricity, it is microscopically neutral.

**Q.58. Where plasma is found?**

Ans. Entire universe is almost of plasma. Plasmas are found in every thing from the sun to quarks. It Is the stuff of stars. Our sun is a 1.5 million kilometer ball of plasma, heated by the nuclear fusion. On earth, it occurs in lightning bolts, flames, auroras and fluorescent lights.

**Q.59. Why the excluded volume (effective volume) is less than molar volume of the gas?**

Ans. Excluded volume ‘b” is the volume occupied by 1 mole of the real gas molecules in highly compressed state. But molar volume V_{m} is the actual volume of one mole of a gas molecules and is the measured container volume which is accessible to each molecule. In a real gas, however, the molecules do occupy some volume. Thus, the excluded volume is less than the molar volume of the gas.

**Q.60. Why the volume correction is done by van der Waal.**

Ans. The real gas molecules do have definite volume. This volume, however, is not negligible when the gas is subjected to high pressure. So van der Waals needed a correction in volume factor in order to make it applicable to the real gases. Thus, the volume available to gas molecules is the volume of the vessel minus the volume of gas molecules.

V_{free} = V_{vessel} – b

The factor b is called the excluded volume which is constant and characteristic of a gas.

Q.61. a gas occupies 100dm^{3} at 283K. what will be its volume at -273^{o}C at constant pressure?

Ans. V_{1} = 100 dm^{3} T_{1} = 283K

V_{2} = ? T_{2} = -273 + 273 = 0 K

**Q.62. What is the density of CH _{4} (g) at 0^{o}C and 1 atmospheric pressure.**

Ans. T = 0^{o}C + 273 k = 273 K ; P 1 atm

Molecular mass of gas ; M = 16 g mol^{-1}

R = 0.0821 dm^{3}atm K^{-1}Mol^{-1}

d =

**Q.63. the amount of pressure which is decreased due to force of attraction is given by a/V ^{2} where ‘a’ is the van der Waal’s constant and V is the volume of vessel?**

Ans. The pressure p’ is proportional to the number of moles of molecules, n, which are hitting the walls of the container. The total force of attraction on any molecules which is hitting a wall is proportional to the concentration of neighboring molecules, n/V , but the number of molecules hitting the wall per unit wall area is also proportional to the concentration, n/V. thus

P’ x or p’ Or

Where a is constant of proportionality and is called van der Waal’s constant if n = 1 mol of a gas, then

P’ Thus, P_{i} = P +

**Q.64. How the various scales of thermometry can be interconverted?**

Ans. There are three scales of thermometry; i.e., centigrade, Fahrenheit and absolute or Kelvin scale. These scales of thermometry can be interconverted as follows:

^{o}C = (G – 32), F = (^{o}C) + 32 and K = ^{o}C +273

**Q.65. Greater the temperature of the gas, closer the straight line of P versus I/V to the pressure axis. Justify it.**

Ans. The volume of the gas increases with the rise in temperature for the same number of moles at the same pressure. Thus inverse of volume, I/V decreases at the same pressure and so I/V will be close to zero when P is close to zero. Therefore, greater the temperature of the gas, closer the straight line of P versus I/V to the pressure axis.

**Q.66. Give two uses of plasma?**

Ans. 1. Plasma can be used to destroy bacteria.

2. Plasma can be used for cleaning and sterilization of food and operation theaters.