JEE Mains · Physics · STD 11 - 11. thermodynamics
One mole of an ideal monoatomic gas is taken along the path \(ABCA\) as shown in the \(PV\) diagram. The maximum temperature attained by the gas along the path \(BC\) is given by
- A \(\frac{{25}}{8}\,\frac{{{P_0}{V_0}}}{R}\)
- B \(\frac{{25}}{4}\,\frac{{{P_0}{V_0}}}{R}\)
- C \(\frac{{25}}{16}\,\frac{{{P_0}{V_0}}}{R}\)
- D \(\frac{{5}}{8}\,\frac{{{P_0}{V_0}}}{R}\)
Answer & Solution
Correct Answer
(A) \(\frac{{25}}{8}\,\frac{{{P_0}{V_0}}}{R}\)
Step-by-step Solution
Detailed explanation
Equation of the \(BC\) \(p = {p_0} - \frac{{2{P_0}}}{{{V_0}}}\left( {V - 2{V_0}} \right)\) using=\(PV=nRT\) Temperature,\(T = \frac{{{P_0}V - \frac{{2{P_0}{V^2}}}{{{V_0}}} + 4{P_0}V}}{{1 \times R}}\) \(\left( {n = 1\,mole\,given} \right)\)…
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