First slide

Change in internal energy

Question

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

Moderate

A

20 J
B

-12 kJ
C

20 kJ
D

- 20 kJ

Solution

$We know, Δ U = n C_{v} Δ T$
$= n (\frac{5 R}{2}) (T_{B} - T_{A}) [for diatomic gas, C_{v} = \frac{5 R}{2}]$
$= \frac{5 n R}{2} (\frac{P_{B} V_{B}}{n R} - \frac{P_{A} V_{A}}{n R}) [∵ P V = n R T]$
$= \frac{5}{2} (P_{B} V_{B} - P_{A} V_{A}) = \frac{5}{2} (2 \times 10^{3} \times 6 - 5 \times 10^{3} \times 4)$
$= \frac{5}{2} (- 8 \times 10^{3}) = - 20 kJ$

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