Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

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The document Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q. 13. A tall cylindrical vessel with gaseous nitrogen is located in a uniform gravitational field in which the free-fall acceleration is equal to g. The temperature of the nitrogen varies along the height h so that its density is the same throughout the volume. Find the temperature gradient dT/dh. 

Solution. 13. Consider a thin layer at a height h and thickness dh. Let p and dp+p be the pressure on the two sides of the layer. The mass of the layer is Sdhp. Equating vertical downward force to the upward force acting on the layer.

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev    (1)

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

That means, temperature of air drops by 34°C at a height of 1 km above bottom.


Q. 14. Suppose the pressure p and the density p of air are related as p/pn = const regardless of height (n is a constant here). Find the corresponding temperature gradient.

Solution. 14.  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev     (1)

But, from    Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev     (2)

We have from gas low   Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Thus,    Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

But,    Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

so,   Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev


Q. 15. Let us assume that air is under standard conditions close to the Earth's surface. Presuming that the temperature and the molar mass of air are independent of height, find the air pressure at the height 5.0 km over the surface and in a mine at the depth 5.0 km below the surface. 

Solution. 15. We have, dp = - p g dh and from gas law  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Integrating, we get

or,  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

(where p0 is the pressure at the surface of the Earth.)

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

[Under standard condition, Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Pressure at a height of 5 atm   Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Pressure in a mine at a depth of 5 km  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev


Q. 16. Assuming the temperature and the molar mass of air, as well as the free-fall acceleration, to be independent of the height, find the difference in heights at which the air densities at the temperature 0°C differ 

(a) e times; (b) by η = 1.0%. 

Solution. 16. We have dp = - pg dh but from gas law  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Thus    Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev at const, temperature 

So,  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Integrating within limits   Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

or,    Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Thus    Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev


Q. 17. An ideal gas of molar mass M is contained in a tall vertical cylindrical vessel whose base area is S and height h. The temperature of the gas is T, its pressure on the bottom base is P0. Assuming the temperature and the free-fall acceleration g to be independent of the height, find the mass of gas in the vessel. 

Solution. 17. From the Barometric formula, we have

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

and from gas law

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

So, at constant temperature from these two Eqs.

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev     (1)

Eq. (1) shows that density varies with height in the same manner as pressure. Let us consider the mass element of the gas contained in the coltimn.

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Hence the sought mass, 

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev


Q. 18. An ideal gas of molar mass M is contained in a very tall vertical cylindrical vessel in the uniform gravitational field in which the free-fall acceleration equals g. Assuming the gas temperature to be the same and equal to T, find the height at which the centre of gravity of the gas is located.

Solution. 18. As the gravitational field is constant the centre of gravity and the centre of mass are same. The location of C.M.

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

But from Barometric formula and gas law  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

So,    Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev


Q. 19. An ideal gas of molar mass /If is located in the uniform gravitational field in which the free-fall acceleration is equal to g. Find the gas pressure as a function of height h, if p = p0  at h = 0, and the temperature varies with height as 

(a) T = T0  (1 — ah); (b) T = T0  (1 + ah),

where a is a positive constant. 

Solution. 19. (a) We know that the variation of pressure with height of a fluid is given by :

dp = - p g dh

But from gas law   Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

From these two Eqs.

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

or,   Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Integrating,  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Hence,   Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

(b) Proceed up to Eq. (1) of part (a), and then put  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev and proceed further in the same fashion to get

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev


Q. 20. A horizontal cylinder closed from one end is rotated with a constant angular velocity ω about a vertical axis passing through the open end of the cylinder. The outside air pressure is equal to P0, the temperature to T, and the molar mass of air to M. Find the air pressure as a function of the distance r from the rotation axis. The molar mass is assumed to be independent of r. 

Solution. 20. Let us consider the mass element of the gas (thin layer) in the cylinder at a distance r from its open end as shown in the figure.

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Using Newton’s second law for the element 

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

So,     Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Thus,    Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev


Q. 21. Under what pressure will carbon dioxide have the density p = 500 g/1 at the temperature T = 300 K? Carry out the calculations both for an ideal and for a Van der Waals gas. 

Solution. 21. For an ideal gas law

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

So,   Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

For Vander Waal gas Eq. 

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

or,   Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev


Q. 22. One mole of nitrogen is contained in a vessel of volume V = 1.00 1. Find:
 (a) the temperature of the nitrogen at which the pressure can be calculated from an ideal gas law with an error η = 10% (as compared with the pressure calculated from the Van der Waals equation of state);
 (b) the gas pressure at this temperature. 

Solution. 22.  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

(The pressure is less for a Vander Waal gas than for an ideal gas)

or,   Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

or,  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

(b) The corresponding pressure is

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev


Q. 23. One mole of a certain gas is contained in a vessel of volume V = 0.250 1. At a temperature T1 = 300 K the gas pressure is p1 = 90 atm, and at a temperature T= 350 K the pressure is p2 = 110 atm. Find the Van der Waals parameters for this gas. 

Solution. 23. Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

So,  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

or.  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Also,  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

or,   Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Using T1 = 300 K, p1 = 90 atms, T2 = 350 K, p2 = 110 atm, V = 0.250 litre 

a = 1.87 atm. litre2/mole2, b = 0.045 litre/mole


Q. 24. Find the isothermal compressibility x of a Van der Waals gas as a function of volume V at temperature T.  

Note. By definition Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Solution. 24. Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

or,   Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev


Q. 25. Making use of the result obtained in the foregoing problem, find at what temperature the isothermal compressibility x of a Van der Waals gas is greater than that of an ideal gas. Examine the case when the molar volume is much greater than the parameter b. 

Solution. 25. For an ideal gas  Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRevIrodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

Now   Irodov Solutions: Equation of The Gas State Processes- 2 Notes | EduRev

If a, b do not vary much with temperature, then the effect at high temperature is clearly determined by b and its effect is repulsive so compressibility is less.

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