An ideal gas,overline{C}_{V} = frac{5}{2} R,i | vedclass

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(B) For an adiabatic process,$\Delta U = w$ (since $q = 0$).$n \overline{C}_{V} (T_2 - T_1) = -P_{ext} (V_2 - V_1)$.Given $V_2 = 2 V_1$,so $V_2 - V_1

Vedclass · Accepted Answer

$274$

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(B) For an adiabatic process,$\Delta U = w$ (since $q = 0$).$n \overline{C}_{V} (T_2 - T_1) = -P_{ext} (V_2 - V_1)$.Given $V_2 = 2 V_1$,so $V_2 - V_1 = V_1$.Using the ideal gas law $V_1 = \frac{n R T_1}{P_1}$,we have $V_2 - V_1 = \frac{n R T_1}{P_1}$.Substituting the values: $n (\frac{5}{2} R) (T_2 - 298) = -1 \ atm 	imes (\frac{n R 	imes 298}{5 \ atm})$.Canceling $n R$ from both sides: $\frac{5}{2} (T_2 - 298) = -\frac{298}{5}$.$T_2 - 298 = -\frac{298 	imes 2}{25} = -\frac{596}{25} = -23.84$.$T_2 = 298 - 23.84 = 274.16 \ K$.The nearest integer is $274 \ K$.

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