First slide

Work done by gas

Question

If a gas expands with temperature according to the relation V = KT^2/3, then what will be work done
when the temperature changes by 30⁰C?

Moderate

A

40R
B

30R
C

20R
D

10R

Solution

Work done, $W = \int p d V = \int \frac{R T}{V} d V$
Since, $V = K T^{2 / 3} \Rightarrow d V = \frac{2}{3} K T^{- V / 3} d T$
Eliminating K, we find $\frac{d V}{V} = \frac{2}{3} \frac{d T}{T}$
Hence, $W = \int_{T_{1}}^{T_{2}} \frac{2}{3} \frac{R T}{T} d T$
$= \frac{2}{3} R (T_{2} - T_{1}) = \frac{2}{3} R (30) = 20 R$

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