Question 470: Thermodynamics & KTG

Question

Two moles of helium gas are taken over the cycle $ABCD$, as shown in the $P$-$T$ diagram.
The net work done on the gas in the cycle $ABCDA$ is:
![](https://imagex.cdn.examgoal.net/kL9IFbFk4ynlJaPv9/nrqT0fOSfEycVZatTX2D1gOuRYozW/f0djQCbWSLYWN9ssqqaMS9/image.png)

Accepted Answer

The net work in the cycle $ABCD$ is
$W = {W_{AB}} + {W_{BC}} + {W_{CD}} + {W_{DA}}$
$ = 400R + 2.303nRT\log {{{P_B}} \over {{P_C}}} + \left( { - 400R} ight) - 414R$
$ = 2.303 	imes 2R 	imes 500\log {{2 	imes {{10}^5}} \over {1 	imes {{10}^5}}} - 414R$
$ = 693.2\,R - 414\,R = 279.2\,R$

Answer

$276$ $R$

Answer

$1076$ $R$

Answer

$1904$ $R$

Answer

zero

Thermodynamics & KTG

Two moles of helium gas are taken over the cycle ABCD, as shown in the P-T diagram. The net work done on the gas in the cycle ABCDA is:

Choose the correct answer: