One mole of a diatomic gas undergoes a process `P = P_(0)//[1 + (V//V_(0)^(3))]` where `P_(0)` and `V_(0)` are constant. The translational kinetic energy of the gas when `V = V_(0)` is given byA. `5 P_(0) V_(0)//4`B. `3 P_(0) V_(0)//4`C. `3 P_(0) V_(0)//2`D. `5 P_(0) V_(0)//2`

One mole of a diatomic gas undergoes a process `P = P_(0)//[1 + (V//V_

1.	One mole of a diatomic gas undergoes a process `P = P_(0)//[1 + (V//V_(0)^(3))]` where `P_(0)` and `V_(0)` are constant. The translational kinetic energy of the gas when `V = V_(0)` is given byA. `5 P_(0) V_(0)//4`B. `3 P_(0) V_(0)//4`C. `3 P_(0) V_(0)//2`D. `5 P_(0) V_(0)//2`
Answer» Correct Answer - B b. `P = (P_(0))/(1 + (V//V_(0))^(3))=(P_(0))/(2)` `T = (P_(0) V_(0))/(2R)` Therefore translational kinetic energy is equal to `(3)/(2) RT = (3R)/(2) (P_(0) V_(0))/(2R) = (3P(0) V_(0))/(4)`

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One mole of a diatomic gas undergoes a process `P = P_(0)//[1 + (V//V_(0)^(3))]` where `P_(0)` and `V_(0)` are constant. The translational kinetic energy of the gas when `V = V_(0)` is given byA. `5 P_(0) V_(0)//4`B. `3 P_(0) V_(0)//4`C. `3 P_(0) V_(0)//2`D. `5 P_(0) V_(0)//2`

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