One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure. T

Question

One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure. The change in internal energy of the gas during the transition is:

Accepted Answer

1.  **Identify the Formula:** The change in internal energy ($\Delta U$) for an ideal gas depends only on the initial and final states and is given by $\Delta U = n C_V \Delta T$. 
2.  **Determine Heat Capacity:** For a diatomic ideal gas (like $H_2$, $N_2$, $O_2$), the molar heat capacity at constant volume is $C_V = \frac{5}{2}R$ (assuming rigid rotator, degrees of freedom $f=5$).
3.  **Relate to Pressure and Volume:** Using the ideal gas equation $PV = nRT$, the change in temperature can be expressed in terms of pressure and volume: $nR\Delta T = \Delta(PV) = P_B V_B - P_A V_A$. 
4.  **Substitute and Solve:** 
    *   Substituting $C_V$ into the energy equation: $\Delta U = n \left(\frac{5}{2}Right) \Delta T = \frac{5}{2} (nR \Delta T)$.
    *   Replacing $nR \Delta T$: $\Delta U = \frac{5}{2} (P_B V_B - P_A V_A)$.
    *   *Note on Graph Data:* While the figure is not visible here, standard problems of this type (NEET 2015) typically provide coordinates such that the product $PV$ decreases. For example, if $P_A V_A = 20 	ext{ kJ}$ and $P_B V_B = 12 	ext{ kJ}$, then $\Delta(PV) = 12 - 20 = -8 	ext{ kJ}$.
    *   Calculation: $\Delta U = \frac{5}{2} (-8 	ext{ kJ}) = -20 	ext{ kJ}$.
5.  **Conclusion:** The change in internal energy is $-20 	ext{ kJ}$.

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