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JEE Mains · Physics · STD 11 - 11. thermodynamics

An ideal gas undergoes a process maintaining relation between pressure \((P)\) and volume \((V)\) as \(P = P_o\left(1 + \left(\dfrac{V_o}{V}\right)^2\right)^{-1}\), where \(P_o\) and \(V_o\) are constants. If two samples \(A\) and \(B\) (two moles each) with initial volumes \(V_o\) and \(3V_o\) respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples, \(T_B - T_A\) is _____. (\(R = \) gas constant)

  1. A \(\dfrac{9P_o V_o}{8R}\)
  2. B \(\dfrac{11P_o V_o}{10R}\)
  3. C \(\dfrac{7P_o V_o}{6R}\)
  4. D \(\dfrac{13P_o V_o}{11R}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\dfrac{11P_o V_o}{10R}\)

Step-by-step Solution

Detailed explanation

The given relation between pressure and volume is: \(P = P_o\left(1 + \left(\dfrac{V_o}{V}\right)^2\right)^{-1}\) For sample A, the initial volume is \(V_A = V_o\). Substituting this into the given relation, the initial pressure \(P_A\) is:…
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