Question

The pressure of a gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa, 50 cc. (a) Calculate the work done by the gas. (b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas?

Answer 1

The pressure of a gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa, 50 cc.

(a) Work done by the gas = - 4.5 J

(b) If no heat is supplied or extracted from the gas, the change in the internal energy of the gas = 4.5 J

Explanation:

Step 1:

Given  

$P_{1}=10 \mathrm{kPa}=10 \times 10^{3} \mathrm{Pa}$

$P_{2}=50 \mathrm{kPa}=50 \times 10^{3} \mathrm{Pa}$

$V_1 = 200\,cc$  

$V_2 = 50\, cc$  

$P_1$ is system of the Initial pressure,  

$P_2$ is system of the Final volume,  

$V_1$ is system of the Initial volume,  

$V_2$ is system of the Final pressure,  

Step 2:

(i) Gas Work done = Pressure (P) × Change in System volume

When pressure also changes we take the sum of the two pressures given. Now,

$P=\frac{1}{2}(10+50) \times 10^{3}=30 \times 10^{3} \mathrm{Pa}$

Work done using the gas system can be done by

$30 \times 10^{3} \times(50-200) \times 10^{-6}=-4.5 J$

Step 3 :

(ii) Since the system does not have heat supplied, ∆Q= 0.

Using the thermodynamics first law

∆Q = − ∆W = 4.5 J