Question:

One mole of an ideal diatomic gas undergoes a process as shown in the figure. The molar specific heat of the gas in the process is: 

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For an ideal diatomic gas, the molar specific heat at constant volume is \( \frac{5R}{2} \). The specific heat for a process involving temperature and volume changes can be derived from the slope of the corresponding graph.
Updated On: Mar 24, 2025
  • \( \frac{3R}{2} \)
  • \( \frac{R}{2} \)
  • \( \frac{5R}{2} \)
  • \( \frac{7R}{2} \)
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The Correct Option is A

Solution and Explanation

The process shown in the figure indicates an isochoric process where the volume of the gas is directly proportional to the inverse of the temperature. For an ideal gas undergoing such a process, the relation between the pressure, volume, and temperature is: \[ V \propto \frac{1}{T} \] This implies that the process follows a path where the slope of \( V \) versus \( \frac{1}{T} \) is constant. For an ideal diatomic gas, the molar specific heat \( C \) is related to the slope of the curve in the diagram. From the relationship for an ideal gas: \[ C = C_V + R \] where \( C_V \) is the molar specific heat at constant volume for a diatomic gas, which is \( \frac{5R}{2} \). Since the process shown is a specific heat process that involves both heat exchange and work, the total molar specific heat will be \( \frac{3R}{2} \). 
Thus, the molar specific heat of the gas in the process is \( \boxed{\frac{3R}{2}} \).

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