One mole of a diatomic gas undergoes a process P= P_ d 1+ ( V V_ d )^ 3 , where P 0 and V…
Physics · NEET · NTA Exams — KINETIC THEORY OF GASES
One mole of a diatomic gas undergoes a process \[P=\frac{P_{d}}{1+\left(\frac{V}{V_{d}}\right)^{3}}\], where P0 and V0 are constants. The translational kinetic energy of the gas when \(V=V_{B}\) is given by
- \(\frac{5 P_{g}^{2} V_{g}}{4}\)
- \(\frac{3 R_{a}^{2} V_{a}}{4}\)
- \(\frac{3 P_{a}^{2} V_{a}}{2}\)
- \(\frac{5 P_{a}^{2} V_{a}}{2}\)
Answer
(B) 3 R_ a ^ 2 V_ a 4
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