In an isobaric process, the work done by a diatomic gas is $10 \mathrm{~J}$, the heat given to the gas will be
In an isobaric process, the work done by a diatomic gas is $10 \mathrm{~J}$, the heat given to the gas will be
- $35 \mathrm{~J}$
- $30 \mathrm{~J}$
- $45 \mathrm{~J}$
- $60 \mathrm{~J}$
Solution
For constant pressure process
$\begin{aligned} & \frac{W}{Q}=\frac{n R \Delta T}{n C_p \Delta T}=\frac{n R \Delta T}{n\left(\frac{f}{2}+1\right) R \Delta T}=\frac{1}{f / 2+1} \\ & \frac{W}{Q}=\frac{1}{\left(\frac{5}{2}+1\right)}=\frac{2}{7} \Rightarrow Q=\frac{7}{2} W=\frac{7}{2} \times 10=35 \mathrm{~J}\end{aligned}$
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Asked in: NEET 2020 (Phase 2)
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