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NEET 2004
Exam
NEET
Year
2004
Subject
Physics
Chapter
Thermodynamics

NEET 2004 · Physics · Thermodynamics

One mole of an ideal gas at an initial temperature of T~ K does 6~...

One mole of an ideal gas at an initial temperature of T KT~ K does 6 R6~ R joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/35 / 3, the final temperature of gas will be:

Options

  1. A(T+2.4)K(T+2.4) K
  2. B(T−2.4)K(T-2.4) K
  3. C(T+4)K(T+4) K
  4. D(T−4)K(T-4) K

Correct answer

D. (T−4)K(T-4) K

Step-by-step solution

We know that work done in adiabatic process is:W=−1V−1(PfVf−PiVi)⇒6R=−153−1(Tf−Ti) Since PV=RT}⇒Tf−Ti=−4⇒Tf=(T−4)K\begin{aligned}& W=\frac{-1}{V-1}\left(P_f V_f-P_i V_i\right) \\ & \Rightarrow \quad 6 R=\frac{-1}{\frac{5}{3}-1}\left(T_f-T_i\right) \\ & \text{ Since }P V=R T\} \\ & \Rightarrow \quad T_f-T_i=-4 \\ & \Rightarrow \quad T_f=(T-4) \mathrm{K}\\ & \end{aligned}
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