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If ' $\Delta Q^{\prime}$ is the amount of heat supplied to 'n' moles of a diatonic gas at constant
pressure, ' $\triangle \mathrm{U}^{\prime}$ is the change in internal energy and ' $\triangle \mathrm{W}^{\prime}$ is the work done, then
$\Delta \mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}$ is
Options:
pressure, ' $\triangle \mathrm{U}^{\prime}$ is the change in internal energy and ' $\triangle \mathrm{W}^{\prime}$ is the work done, then
$\Delta \mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}$ is
Solution:
2898 Upvotes
Verified Answer
The correct answer is:
$2: 5: 7$
We know,
$$
\begin{array}{l}
\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T} \\
\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T} \\
\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T} \\
\Delta \mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}=\mathrm{R}: \mathrm{C}_{\mathrm{v}}: \mathrm{C}_{\mathrm{p}}
\end{array}
$$
For a diatomic gas $f=5$
$$
\begin{aligned}
\therefore \quad & C_{v}=\frac{f}{2} R=\frac{5}{2} R \\
& C_{p}=C_{v}+R=\frac{7}{2} R \\
\therefore \quad R: C_{v}: C_{p}=1: \frac{5}{2}: \frac{7}{2}=2: 5: 7
\end{aligned}
$$
$$
\begin{array}{l}
\Delta \mathrm{Q}=\mathrm{nC}_{\mathrm{p}} \Delta \mathrm{T} \\
\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T} \\
\mathrm{W}=\mathrm{P} \Delta \mathrm{V}=\mathrm{nR} \Delta \mathrm{T} \\
\Delta \mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}=\mathrm{R}: \mathrm{C}_{\mathrm{v}}: \mathrm{C}_{\mathrm{p}}
\end{array}
$$
For a diatomic gas $f=5$
$$
\begin{aligned}
\therefore \quad & C_{v}=\frac{f}{2} R=\frac{5}{2} R \\
& C_{p}=C_{v}+R=\frac{7}{2} R \\
\therefore \quad R: C_{v}: C_{p}=1: \frac{5}{2}: \frac{7}{2}=2: 5: 7
\end{aligned}
$$
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