JEE Mains · Physics · STD 11 - 11. thermodynamics
10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from \(P _1\) to \(P _2\) is \(\alpha\) Joule \(\left( P _1=21.7 Pa\right.\) and \(\left.P _2=30 Pa, C _{ v }=21 J / K . mol , R=8.3 J / mol . K \right)\). The value of \(\alpha\) is _________ .
- A 24
- B 15
- C 21
- D 28
Answer & Solution
Correct Answer
(C) 21
Step-by-step Solution
Detailed explanation
\(\Delta Q=nC_{v}\Delta T\) (isochoric) \(=\frac{C_{v}}{R}\cdot nR\Delta T=\frac{C_{v}}{R}(P_{2}-P_{1})V\) \(=\frac{21}{8.3}\times(30-21.7)\times1=21J\)
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