JEE Mains · Physics · STD 11 - 11. thermodynamics
The temperature of \(3.00\, {mol}\) of an ideal diatomic gas is increased by \(40.0^{\circ} {C}\) without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is \(\frac{{x}}{10} .\) Then the value of \({x}\) (round off to the nearest integer) is ..... . \(\left(\right.\) Given \(\left.{R}=8.31\, {J} {mol}^{-1} {K}^{-1}\right)\)
- A \(25\)
- B \(2.5\)
- C \(125\)
- D \(250\)
Answer & Solution
Correct Answer
(A) \(25\)
Step-by-step Solution
Detailed explanation
Pressure is not changing \(\Rightarrow\) isobaric process \(\Rightarrow \Delta {U}={n} {C}_{{v}} \Delta {T}=\frac{5 {nR} \Delta {T}}{2}\) and \({W}={n} {R} \Delta {T}\) \(\frac{\Delta {U}}{{W}}=\frac{5}{2}=\frac{{x}}{10} \Rightarrow {x}=25.00\)
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