## Saturday, February 17, 2007

### 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 TVγ–1 = constant.
Therefore we have 300 V0.5 = T(V/16)0.5. Note that the temperature is to be substituted in Kelvin. The final temperature is given by T = 300×160.5 = 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 PVγ = constant.
Differentiating, P γVγ–1 dV + VγdP = 0
Slope of adiabatic curve = dP/dV = (– γPVγ–1)/Vγ = – 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 PVγ = constant so that P×8005/3 = P'×1005/3 from which P' = P×85/3 = 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).