Two moles of an ideal monoatomic gas at 27^{c | vedclass

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(B) For an adiabatic process,the work done by the gas is given by $W = \frac{\mu R(T_1 - T_2)}{\gamma - 1}$.Using the adiabatic relation $T_1 V_1^{\ga

Vedclass · Accepted Answer

$2767.23$

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(B) For an adiabatic process,the work done by the gas is given by $W = \frac{\mu R(T_1 - T_2)}{\gamma - 1}$.Using the adiabatic relation $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$,we have $T_2 = T_1 \left(\frac{V_1}{V_2}ight)^{\gamma-1}$.Substituting this into the work formula: $W = \frac{\mu R T_1}{\gamma - 1} \left[ 1 - \left(\frac{V_1}{V_2}ight)^{\gamma-1} ight]$.Given: $\mu = 2 	ext{ mol}$,$T_1 = 27 + 273 = 300 	ext{ K}$,$V_1 = V$,$V_2 = 2V$,$\gamma = 5/3$,$R = 8.31 	ext{ J/mol K}$.$W = \frac{2 	imes 8.31 	imes 300}{(5/3 - 1)} \left[ 1 - \left(\frac{V}{2V}ight)^{5/3 - 1} ight]$.$W = \frac{4986}{2/3} \left[ 1 - (1/2)^{2/3} ight]$.$W = 7479 	imes [1 - 0.62996] = 7479 	imes 0.37004 \approx 2767.23 	ext{ J}$.

Two moles of an ideal monoatomic gas at $27^{\circ}C$ occupies a volume of $V$. If the gas is expanded adiabatically to the volume $2V$,then the work done by the gas will be ....... $J$ $[\gamma = 5/3, R = 8.31 \text{ J/mol K}]$

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