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:

Question

One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure, <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-yuwjldrmr3pul5bu.jpg" /> The change in internal energy of the gas during the transition is: (A) 20kJ (B) -20kJ (C) 20J (D) -12kJ

Tardigrade Admin · Accepted Answer

Δ U=nCvΔ T T=(P V/n R) Δ T=(P2 V2 - P1 V2/n R) Δ U=(n R/&gamma; - 1) ((P2 V2 - P1 V1/n R)) =(P2 V2 - P1 V1/&gamma; - 1) =-(8 × 103/(2)5)=-20 kJ

Q. 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:

$20 k J$

$- 20 k J$

$20 J$

$- 12 k J$

$Δ U = n C_{v} Δ T$
$T = \frac{P V}{n R}$
$Δ T = \frac{P _{2} V _{2} - P _{1} V _{2}}{n R}$
$Δ U = \frac{n R}{γ - 1} (\frac{P _{2} V _{2} - P _{1} V _{1}}{n R})$
$= \frac{P _{2} V _{2} - P _{1} V _{1}}{γ - 1}$
$= - \frac{8 \times 1 0 ^{3}}{\frac{2}{5}} = - 20 k J$

Q. 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:

20kJ

−20kJ

20J

−12kJ