- Home
- Standard 11
- Physics
11.Thermodynamics
hard
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 ............$\;kJ$
A
$20$
B
$-20$
C
$0.02$
D
$-12$
(AIPMT-2015)
Solution
We know, $\Delta U = n{C_v}\Delta T$
$ = n\left( {\frac{{5R}}{2}} \right)\left( {{T_B} – {T_A}} \right)$ $[for\,diatomic\,gas,{C_v} = \frac{{5R}}{2}]$
$ = \frac{{5nR}}{2}\left( {\frac{{{P_B}{V_B}}}{{nR}} – \frac{{{P_A}{V_A}}}{{nR}}} \right)$
$\left[ {PV = nRT} \right]$
$ = \frac{5}{2}\left( {{P_B}{V_B} – {P_A}{V_A}} \right)$
$ = \frac{5}{2}\left( {2 \times {{10}^3} \times 6 – 5 \times {{10}^3} \times 4} \right)$
$ = \frac{5}{2}\left( { – 8 \times {{10}^3}} \right) = – 20\,kJ$
Standard 11
Physics
Similar Questions
hard
