One mole of an ideal gas is allowed to expand | vedclass

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(B) For an adiabatic process,the change in internal energy $\Delta U$ is equal to the work done $w$ on the system. According to the first law of therm

Vedclass · Accepted Answer

$150$

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(B) For an adiabatic process,the change in internal energy $\Delta U$ is equal to the work done $w$ on the system. According to the first law of thermodynamics,$\Delta U = q + w$. Since the process is adiabatic,$q = 0$,so $\Delta U = w$.Given that the gas expands,work is done by the system,so $w = -3000\, J$.Using the relation $\Delta U = nC_V\Delta T$,where $n = 1\, mol$ and $C_V = 20\, J/(mol\cdot K)$:$-3000\, J = 1\, mol 	imes 20\, J/(mol\cdot K) 	imes (T_f - T_i)$.Initial temperature $T_i = 27 + 273 = 300\, K$.$-3000 = 20 	imes (T_f - 300)$.$-150 = T_f - 300$.$T_f = 300 - 150 = 150\, K$.

One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27\,^{\circ}C$. If the work done during the process is $3\, kJ$,then the final temperature of the gas is $...\, K$.

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