The percentage of heat supplied to a diatomic | vedclass

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(C) For an isobaric process, the pressure remains constant $(P = 	ext{constant})$.From the first law of thermodynamics, $\Delta Q = \Delta U + \Delta

Vedclass · Accepted Answer

$28.6$

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(C) For an isobaric process, the pressure remains constant $(P = 	ext{constant})$.From the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$.For a diatomic gas, the molar heat capacities are $C_p = \frac{7}{2}R$ and $C_v = \frac{5}{2}R$.The heat supplied is $\Delta Q = n C_p \Delta T$.The work done is $\Delta W = n R \Delta T$.The fraction of heat converted into work is $\frac{\Delta W}{\Delta Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p} = \frac{R}{\frac{7}{2}R} = \frac{2}{7}$.Converting this to a percentage: $\frac{2}{7} 	imes 100 \approx 28.6 \%$.Therefore, the percentage of heat converted into work is $28.6 \%$.

The percentage of heat supplied to a diatomic ideal gas that is converted into work in an isobaric process is

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