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TS EAMCET · Physics · Thermodynamics

A polyatomic gas follows a law \(T^2 V^\alpha=\) constant. Find \(\alpha\) for which the heat exchange of gas in the process becomes zero.

  1. A \(\alpha=\frac{3}{2}\)
  2. B \(\alpha=\frac{2}{3}\)
  3. C \(\alpha=\frac{4}{3}\)
  4. D \(\alpha=\frac{3}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\alpha=\frac{2}{3}\)

Step-by-step Solution

Detailed explanation

The process in which there is no heat exchange of gas, it is called adiabatic process. In this process, \(T V^{\gamma-1}=k\) [constant \(]\) Now, \(T^2 V^{2(\gamma-1)}=k^2= constant\)....(i) Given that, \(T^2 V^\alpha=\) constant.....(ii) Comparing Eqs. (i) and (ii), we get…