Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

20 kJ

-20 kJ

20 J

-12 kJ

Step-by-Step Solution

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 $\Delta U = n C_V \Delta T$ .
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 $f=5$ ).
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: $nR\Delta T = \Delta(PV) = P_B V_B - P_A V_A$ .
Substitute and Solve: Substituting $C_V$ into the energy equation: $\Delta U = n \left(\frac{5}{2}R\right) \Delta T = \frac{5}{2} (nR \Delta T)$ . 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 $P_A V_A = 20 \text{ kJ}$ and $P_B V_B = 12 \text{ kJ}$ , then $\Delta(PV) = 12 - 20 = -8 \text{ kJ}$ .
Calculation: $\Delta U = \frac{5}{2} (-8 \text{ kJ}) = -20 \text{ kJ}$ .

Conclusion: The change in internal energy is $-20 \text{ kJ}$ .

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