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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?
Solution
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
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