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An insulating cylinder contains 4 moles of an ideal diatomic gas. When a heat $\mathrm{Q}$ is supplied, to it, 2 moles of the gas molecules dissociate. If the temperature of the gas remains constant, then the value of $Q$ is
$$
\text { ( } \mathbf{R} \text { - universal gas constant) }
$$
Options:
$$
\text { ( } \mathbf{R} \text { - universal gas constant) }
$$
Solution:
1967 Upvotes
Verified Answer
The correct answer is:
RT
Heat supplied " $Q$ " is given by:
$$
\begin{aligned}
& \mathrm{Q}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}} \\
& \mathrm{U}_{\mathrm{i}}=\mathrm{Initial}_{\text {Internal Energy }} \\
& \mathrm{U}_{\mathrm{f}}=\text { Final Internal Energy }
\end{aligned}
$$
Initially cylinder contains 4 moles of diatomic gas,
$$
\begin{aligned}
& \mathrm{U}_{\mathrm{i}}=\mathrm{n}\left(\frac{\mathrm{f}}{2} \mathrm{RT}\right) \\
& \mathrm{U}=4[-\mathrm{RT}]=10 \mathrm{RT}
\end{aligned}
$$
2 moles of diatomic gas dissociate into monoatomic gas. Hence there are 4 moles of monoatomic gas and 2 moles of diatomic gas.
Final Internal Energy is given by:
$$
\begin{aligned}
& U_f=n\left(\frac{f}{2} R T\right)+n^{\prime}\left(\frac{f^{\prime}}{2}-R T\right) \\
& =4\left(\frac{3}{2} R T\right)+2\left(\frac{5}{2} R T\right) \\
& =11 R T
\end{aligned}
$$
Thus Heat supplied is,
$$
\begin{aligned}
& \mathrm{Q}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}} \\
& =11 \mathrm{RT}-10 \mathrm{RT}=\mathrm{RT} \\
& \mathrm{Q}=\mathrm{RT} .
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{Q}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}} \\
& \mathrm{U}_{\mathrm{i}}=\mathrm{Initial}_{\text {Internal Energy }} \\
& \mathrm{U}_{\mathrm{f}}=\text { Final Internal Energy }
\end{aligned}
$$
Initially cylinder contains 4 moles of diatomic gas,
$$
\begin{aligned}
& \mathrm{U}_{\mathrm{i}}=\mathrm{n}\left(\frac{\mathrm{f}}{2} \mathrm{RT}\right) \\
& \mathrm{U}=4[-\mathrm{RT}]=10 \mathrm{RT}
\end{aligned}
$$
2 moles of diatomic gas dissociate into monoatomic gas. Hence there are 4 moles of monoatomic gas and 2 moles of diatomic gas.
Final Internal Energy is given by:
$$
\begin{aligned}
& U_f=n\left(\frac{f}{2} R T\right)+n^{\prime}\left(\frac{f^{\prime}}{2}-R T\right) \\
& =4\left(\frac{3}{2} R T\right)+2\left(\frac{5}{2} R T\right) \\
& =11 R T
\end{aligned}
$$
Thus Heat supplied is,
$$
\begin{aligned}
& \mathrm{Q}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}} \\
& =11 \mathrm{RT}-10 \mathrm{RT}=\mathrm{RT} \\
& \mathrm{Q}=\mathrm{RT} .
\end{aligned}
$$
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