Two moles of a gas at a temperature of 327^{c | vedclass

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(A) Given: Number of moles $n = 2$, Initial temperature $T_1 = 327 + 273 = 600 \ K$, Ratio of specific heats $\gamma = \frac{4}{3}$.Volume increases b

Vedclass · Accepted Answer

$14.94$

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(A) Given: Number of moles $n = 2$, Initial temperature $T_1 = 327 + 273 = 600 \ K$, Ratio of specific heats $\gamma = \frac{4}{3}$.Volume increases by $700 \%$, so $V_2 = V_1 + 7V_1 = 8V_1$.For an adiabatic process, $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.$T_2 = T_1 \left( \frac{V_1}{V_2} ight)^{\gamma-1} = 600 \left( \frac{1}{8} ight)^{\frac{4}{3}-1} = 600 \left( \frac{1}{8} ight)^{1/3} = 600 	imes \frac{1}{2} = 300 \ K$.Work done in an adiabatic process is $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.$W = \frac{2 	imes 8.3 	imes (600 - 300)}{\frac{4}{3} - 1} = \frac{2 	imes 8.3 	imes 300}{1/3} = 2 	imes 8.3 	imes 300 	imes 3 = 14940 \ J = 14.94 \ kJ$.

Two moles of a gas at a temperature of $327^{\circ} C$ expands adiabatically such that its volume increases by $700 \%$. If the ratio of the specific heat capacities of the gas is $\frac{4}{3}$, then the work done by the gas is (Universal gas constant $= 8.3 \ J \ mol^{-1} \ K^{-1}$) (in $kJ$)

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