A monoatomic gas (gamma = 5/3) expands at con | vedclass

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(A) For an isobaric process, the heat supplied is $dQ = nC_p dT$. The change in internal energy is $dU = nC_v dT$. The work done is $dW = PdV = nR dT$

Vedclass · Accepted Answer

$60\%, 40\%$

Vedclass · Answer

(A) For an isobaric process, the heat supplied is $dQ = nC_p dT$. The change in internal energy is $dU = nC_v dT$. The work done is $dW = PdV = nR dT$. The fraction of heat converted to internal energy is $\frac{dU}{dQ} = \frac{nC_v dT}{nC_p dT} = \frac{C_v}{C_p} = \frac{1}{\gamma}$. Given $\gamma = 5/3$, the fraction is $\frac{1}{5/3} = 3/5 = 0.6$ or $60\%$. The fraction of heat converted to work is $\frac{dW}{dQ} = \frac{nR dT}{nC_p dT} = \frac{R}{C_p} = \frac{C_p - C_v}{C_p} = 1 - \frac{1}{\gamma} = 1 - 0.6 = 0.4$ or $40\%$. Thus, the percentages are $60\%$ and $40\%$ respectively.

$A$ monoatomic gas $(\gamma = 5/3)$ expands at constant pressure when heat is supplied. What percentage of the heat supplied is converted into internal energy and work, respectively?

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