First slide

Change in internal energy

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

Moderate

A

20 kJ
B

-20 kJ
C

20 J
D

-12 kJ

Solution

At point A : PV = nRT_A $\Rightarrow 5 \times {10^3} \times 4 = nR{T_A}$
$20 \times {10^3} = nR{T_A}$
At point B : PV = nRT_B $\Rightarrow 2 \times {10^3} \times 6 = nR{T_B}$
$12 \times {10^3} = nR{T_B}$
$\Delta U = n{C_V}\Delta T = n\frac{{5R}}{2}\left( {{T_B} - {T_A}} \right)$
$= \frac{5}{2}\left( {nR{T_B} - nR{T_A}} \right) = \frac{5}{2}\left( {12 \times {{10}^3} - 20 \times {{10}^3}} \right)$
$= \frac{5}{2}\left( { - 8 \times {{10}^3}} \right) = - 20 \times {10^3} = - 20KJ$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App