At $N.T.P.$ one mole of diatomic gas is compressed adiabatically to half of its volume $\gamma = 1.41$. The work done on gas will be ....... $J$
Medium
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(c)${T_2} = {T_1}{\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}} = 273{(2)^{0.41}}$$ = 273 \times 1.328 = 363K$
$W = \frac{{R({T_1} - {T_2})}}{{\gamma - 1}} = \frac{{8.31(273 - 363)}}{{1.41 - 1}}$$ = - \,1824$
==> $|W| \approx 1815 J$
$W = \frac{{R({T_1} - {T_2})}}{{\gamma - 1}} = \frac{{8.31(273 - 363)}}{{1.41 - 1}}$$ = - \,1824$
==> $|W| \approx 1815 J$
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