When One mole ideal gas is compressed adiabatically at $27 ° C$ . Its temperature becomes $102 ° C$ . If the work done in this process is NR joules, find the value of N. ( $γ$ = 1.5)

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answer is 150.

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Detailed Solution

Given: $T_{1} = 27^{\circ} C, T_{2} = 102^{\circ} C ; V = 1.5$

Work done in Adiabatic process is given by
$\begin{array}{l} W = - \frac{n R}{V - 1} [T_{1} - T_{2}] = \frac{n R}{V - 1} [T_{2} - T_{1}] \\ W = \frac{1 \times R}{1.5 - 1} [102 - 27] = 150 R J \end{array}$

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