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Q.

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?

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a

40R

b

30R

c

20R

d

10R

answer is C.

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Detailed 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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If a gas expands with temperature according to the relation V = KT2/3, then what will be work donewhen the temperature changes by 300C?