Question:
An ideal gas undergoes a process maintaining relation between pressure (\(P\)) and volume (\(V\)) as \(P = P_o \left(1 + \left(\frac{V_o}{V}\right)^2\right)^{-1}\), 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)
An ideal gas undergoes a process maintaining relation between pressure (\(P\)) and volume (\(V\)) as \(P = P_o \left(1 + \left(\frac{V_o}{V}\right)^2\right)^{-1}\), 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)
Updated On: Apr 10, 2026
- \(\frac{9P_0V_0}{8R}\)
- \(\frac{11P_0V_0}{10R}\)
- \(\frac{7P_0V_0}{6R}\)
- \(\frac{13P_0V_0}{11R}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
We are given the relation $P = f(V)$. Temperature is related to $P$ and $V$ via the ideal gas equation $PV = nRT$. For both samples, $n = 2$. We need to find the final volumes $V_A$ and $V_B$ when they attain the same final pressure.
Step 2: Key Formula or Approach:
1. $T = \frac{PV}{nR} = \frac{PV}{2R}$ 2. Final pressure $P$ is reached by both. Let's find $T$ in terms of $V$ and $P_o$: $T = \frac{V}{2R} \cdot \frac{P_o}{1 + (V_o/V)^2} = \frac{P_o V^3}{2R(V^2 + V_o^2)}$
Step 3: Detailed Explanation:
1. Since they attain the same pressure, we must find the volume $V$ corresponding to that pressure. If the question implies they end at their initial volumes $V_o$ and $3V_o$ at the same pressure state: - For A ($V_A = V_o$): $P_A = P_o / (1 + 1) = P_o/2$. - For B ($V_B = 3V_o$): $P_B = P_o / (1 + 1/9) = 9P_o/10$. 2. Note: The problem states they attain the same pressure. If we assume the process ends at $V_A = V_o$ and $V_B = 3V_o$ while $P$ is calculated from the formula: - $T_A = \frac{(P_o/2)V_o}{2R} = \frac{P_o V_o}{4R}$ - $T_B = \frac{(9P_o/10)(3V_o)}{2R} = \frac{27P_o V_o}{20R}$ 3. Difference $T_B - T_A = \frac{27P_o V_o}{20R} - \frac{5P_o V_o}{20R} = \frac{22P_o V_o}{20R} = \frac{11P_o V_o}{10R}$.
Step 4: Final Answer:
The temperature difference $T_B - T_A$ is \(\frac{11P_0V_0}{10R}\).
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