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(A) For an isobaric process,the first law of thermodynamics is given by $\Delta Q = \Delta U + \Delta W$.We know that $\Delta Q = n C_p \Delta T$,$\De

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$40$

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(A) For an isobaric process,the first law of thermodynamics is given by $\Delta Q = \Delta U + \Delta W$.We know that $\Delta Q = n C_p \Delta T$,$\Delta U = n C_v \Delta T$,and $\Delta W = n R \Delta T$.The fraction of heat used for external work is $\frac{\Delta W}{\Delta Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p}$.Since $C_p = \frac{\gamma R}{\gamma - 1}$,we have $\frac{\Delta W}{\Delta Q} = \frac{R}{\frac{\gamma R}{\gamma - 1}} = \frac{\gamma - 1}{\gamma}$.Given $\gamma = 5/3$,the fraction is $\frac{5/3 - 1}{5/3} = \frac{2/3}{5/3} = \frac{2}{5} = 0.4$.Converting to percentage,$0.4 	imes 100 = 40\%$.

When an ideal gas is heated at constant pressure,what percentage of the heat supplied is used to perform external work? $(\gamma = 5/3)$

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