ThermodynamicsHard
Question
An ideal gaseous sample at initial state (Po, Vo, To) is allowed to expand to volume 2Vo using two different processes. For the first process, the equation of process is 2PV2 = K1 and for the second process, the equation of the process is PV = K2.
Options
A.Magnitude of work done in the first process will be greater than that in the second process.
B.Magnitude of work done in the second process will be greater than that in the first process.
C.Work done in both the processes cannot be compared without knowing the relation between K1 and K2.
D.First process is impossible.
Solution
$w_{1} = - \int_{V_{1}}^{V_{2}}{P.dV} = - \int_{V_{0}}^{2V_{0}}\frac{K_{1}}{2V^{2}}.dV = + \frac{K_{1}}{2}\left( \frac{1}{2V_{0}} - \frac{1}{V_{0}} \right) = - \frac{K_{1}}{4V_{0}}$
$= - \frac{2P_{0}V_{0}^{2}}{4V_{0}} = - \frac{P_{0}V_{0}}{2} = - 0.5\text{ }\text{P}_{0}V_{0} $$$w_{2} = - \int_{V_{1}}^{V_{2}}{P.dV} = - \int_{V_{0}}^{2V_{0}}{\frac{K_{1}}{V}dV} = - K_{2}\ln\frac{2V_{0}}{V_{0}} = - P_{0}V_{0} \times 0.7$$
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