One mole of an ideal gas at an initial temper | vedclass

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(D) For an adiabatic process,the work done $W$ by $n$ moles of an ideal gas is given by the formula:$W = \frac{nR(T_i - T_f)}{\gamma - 1}$Given:$n = 1

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$(T - 4) \ K$

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(D) For an adiabatic process,the work done $W$ by $n$ moles of an ideal gas is given by the formula:$W = \frac{nR(T_i - T_f)}{\gamma - 1}$Given:$n = 1 	ext{ mole}$$T_i = T \ K$$W = 6R \ J$$\gamma = 5/3$Substituting these values into the formula:$6R = \frac{1 \cdot R(T - T_f)}{(5/3 - 1)}$$6R = \frac{R(T - T_f)}{2/3}$$6R = \frac{3R(T - T_f)}{2}$Dividing both sides by $R$ and solving for $T_f$:$6 = 1.5(T - T_f)$$T - T_f = 6 / 1.5 = 4$$T_f = T - 4$Therefore,the final temperature of the gas is $(T - 4) \ K$.

One mole of an ideal gas at an initial temperature of $T \ K$ does $6R \ J$ of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5/3$,the final temperature of the gas will be:

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