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

Initial pressure of the system, P₁ = 10 kPa = 10 × 10³ Pa

Final pressure of the system, P₂ = 50 kPa = 50 × 10³ Pa

Initial volume of the system, V₁ = 200 cc

Final volume of the system, V₂ = 50 cc

(i) Work done on the gas = Pressure × Change in volume of the system

Since pressure is also changing, we take the average of the given two pressures.

Now,

\[P=\left( \frac{1}{2} \right)\left( 10 + 50 \right) \times  {10}^3\]

\[=30 \times  {10}^3\]Pa

Work done by the system of gas can be given by

\[30 \times  {10}^3  \times \left( 50 - 200 \right) \times  {10}^{- 6} \]

\[ =  - 4 . 5  J\]

(ii) Since no heat is supplied to the system, ∆Q = 0.

Using the first law of thermodynamics, we get

∆U = − ∆W = 4.5 J