MHT CET · Chemistry · Thermodynamics (C)
When 1 mole of gas is heated at constant volume, the temperature rises form \(273 \mathrm{~K}\) to \(546 \mathrm{~K}\). If heat supplied to the gas is \(\mathrm{xJ}\), then find the correct statement from following.
- A \(\mathrm{Q}=\Delta \mathrm{U}=\mathrm{x}, \mathrm{W}=0\)
- B \(\mathrm{Q}=\mathrm{W}=\mathrm{x} J, \Delta \mathrm{V}=0\)
- C \(\Delta \mathrm{V}=0, \mathrm{Q}=\mathrm{W}=-\mathrm{x} J\)
- D \(\mathrm{Q}=-\mathrm{W}=\mathrm{x} J, \Delta \mathrm{V}=0\)
Answer & Solution
Correct Answer
(A) \(\mathrm{Q}=\Delta \mathrm{U}=\mathrm{x}, \mathrm{W}=0\)
Step-by-step Solution
Detailed explanation
\(
\mathrm{W}=- \text { Pext. } \Delta \mathrm{V}
\)
Due to constant volume \(\Delta \mathrm{V}=0, \mathrm{~W}=0\)
\(
q=x J
\)
First law of thermodynamics,
\(
\begin{aligned}
& \Delta \mathrm{U}=\mathrm{q}+\mathrm{w} \\
& \Delta \mathrm{U}=\mathrm{xJ}=\mathrm{q}
\end{aligned}
\)
\mathrm{W}=- \text { Pext. } \Delta \mathrm{V}
\)
Due to constant volume \(\Delta \mathrm{V}=0, \mathrm{~W}=0\)
\(
q=x J
\)
First law of thermodynamics,
\(
\begin{aligned}
& \Delta \mathrm{U}=\mathrm{q}+\mathrm{w} \\
& \Delta \mathrm{U}=\mathrm{xJ}=\mathrm{q}
\end{aligned}
\)
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