Two Questions on Specific Heat Capacity of a Gas

Questions similar to the following one are often found in Medical and Engineering entrance examination question papers:

One mole of an ideal mono atomic gas is mixed with two moles of an ideal diatomic gas. The ratio of specific heats of the mixture is

(a) 1.5 (b) 1.4 (c) 10/6 (d) 15/11 (e) 19/13

You should remember that the values of molar specific heats at constant volume C_v) for mono atomic and diatomic gases are respectively (3/2)R and (5/2)R where R is universal gas constant. The values of molar specific heat at constant pressure C_p) are therefore (5/2)R and (7/2)R respectively, in accordance with Meyer’s relation [C_p – C_v =R].

Therefore, C_v of the mixture = [1×(3/2)R + 2×(5/2)R]/(1+2) = (13/6)R

C_p of the mixture = C_v + R = (19/6)R.

Ratio of specific heats of the mixture, γ = C_p/C_v = 19/13.

[Generally, if n₁ moles of a gas having ratio of specific heats γ₁ is mixed with n₂ moles of a gas having ratio of specific heats γ₂, the ratio of specific heats of the mixture is given by the relation, (n₁+ n₂)/(γ– 1) = n₁/( γ₁ –1) + n₂/( γ₂ –1). You can easily arrive at this result].

If one mole of an ideal mono atomic gas is mixed with one mole of an ideal diatomic gas, the ratio of specific heats of the mixture is 1.5. As an exercise, check this.

Now consider the following MCQ:

The following sets of experimental values of C_v and C_p of a given sample of gas were reported by five groups of students. The unit used is calorie mole^–1 K^–1. Which set gives the most reliable values?

(a) C_v = 3, C_p = 4.5 (b) C_v = 2, C_p = 4 (c) C_v = 3, C_p = 4.9 (d) C_v = 2.5, C_p = 4.5 (e) C_v = 3, C_p = 4.2

Since the minimum value of C_v is (3/2)R which is the value for a mono atomic gas, when you express it in calorie mole^–1 K^–1, the minimum value is approximately 3. [R = 8.3 J mole^–1 K^–1 = 2 calorie mole^–1 K^–1, approximately]. Options (b) and (d) are therefore not acceptable. Out of the remaining three options, (c) is the most reliable since C_p – C_v = R, which should be 2 calorie mole^–1 K^–1 very nearly.

Questions on Isothermal and Adiabatic Changes

You can expect questions involving isothermal and adiabatic processes in most entrance examinations. Here is a typical question:
A gas at a temperature of 27°C inside a container is suddenly compressed to one sixteenths of its initial volume. The temperature of the gas immediately after the compression is (Ratio of specific heats of the gas, γ = 1.5)
(a) 19200 K (b) 1200°C (c) 927°C (d) 108°C (e) 19200°C
This is an adiabatic change since the compression is sudden so that the volume(V) and the temperature (T) are related as TV^γ–1 = constant.
Therefore we have 300 V^0.5 = T(V/16)^0.5. Note that the temperature is to be substituted in Kelvin. The final temperature is given by T = 300×16^0.5 = 1200 K = 927°C.
The following MCQ is meant for testing your understanding of the work done in thermodynamic processes:

Starting from the same initial conditions an ideal gas expands from volume V₁ to volume V₂ in three different ways: (i) Adiabatically, doing work W₁. (ii) Isothermally, doing work W₂. (iii) Isobarically, doing work W₃. Then,

(a) W₁ = W₂ = W₃ (b) W₁ > W₂ > W₃ (c) W₃ > W₂ > W₁ (d) W₂ > W₁ > W₃ (e) W₁ > W₃ > W₂

Since the work done is the area under the corresponding curve in a PV diagram, you can easily verify that the work done is the largest in the isobaric case since it is a straight line parallel to the volume axis. The adiabatic curve is the steepest one so that the area under it is the smallest. The correct option therefore is (c).
You can easily show that the slope of the adiabatic curve is γ times the slope ofthe isothermal curve as follows:
In the case of an adiabatic change, the pressure and volume are related as PV^γ = constant.
Differentiating, P γV^γ–1 dV + V^γdP = 0
Slope of adiabatic curve = dP/dV = (– γPV^γ–1)/V^γ = – γP/V
In the case of an isothermal change, the pressure and volume are related as PV= constant.
Differentiating, PdV + VdP = 0.
Slope of isothermal curve = dP/dV = – P/V.
This show that the slope of the adiabatic curve is γ times the slope ofthe isothermal curve.
The following MCQ also pertains to adiabatic compression:
A gas having a volume of 800 cm³ is suddenly compressed to 100cm³. If the initial pressure is P, the final pressure is (γ = 5/3)
(a) P/32 (b) 24P (c) 32P (d) 8P (e) 16P
We have PV^γ = constant so that P×800^5/3 = P'×100^5/3 from which P' = P×8^5/3 = P×2⁵ = 32P.

Now, see this simple question:
An ideal gas expands isothermally from volume V₁ to volume V₂. It is then compressed to the original volume V₁ adiabatically. The initial pressure is P₁, final pressure is P₂ and the net work done by the gas during the entire process is W. Then
(a) P₁ = P₂, W>0 (b) P₁> P₂, W>0 (c) P₂ > P₁, W>0 (d) P₂ > P₁, W=0 (e) P₂ > P₁, W<0

The adiabatic compression will increase the temperature of the gas so that the final pressure (P₂) when the volume is restored to the value V₁ is greater than the initial pressure P₁. Since the pressure is greater during the adiabatic compression, more work has to be done on the gas. The work done on the gas is thus greater than the work done by the gas. In other words, the net work done by the gas during the entire process is negative. So, the correct option is (e).