JEE Mains · Physics · STD 11 - 11. thermodynamics
'\(n\)' moles of an ideal gas undergoes a process \(A \rightarrow B\) as shown in the figure. The maximum temperature of the gas during the process will be
- A \(\frac{{9{P_0}{V_0}}}{{2nR}}\)
- B \(\;\frac{{9{P_0}{V_0}}}{{nR}}\)
- C \(\;\frac{{9{P_0}{V_0}}}{{4nR}}\)
- D \(\;\frac{{3{P_0}{V_0}}}{{2nR}}\)
Answer & Solution
Correct Answer
(C) \(\;\frac{{9{P_0}{V_0}}}{{4nR}}\)
Step-by-step Solution
Detailed explanation
The euatione for the line is \(P = \frac{{ - {P_0}}}{{{V_0}}}V + 3P\) \([slope = \frac{{ - {P_0}}}{{{V_0}}},c = 3{P_0}]\) \(P{V_0} + {P_0}V = 3{P_0}{V_0}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\) \(But\,\,\,\,\,\,\,PV = nRT\,\)…
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