- Home
- Standard 11
- Physics
11.Thermodynamics
medium
Two moles of helium gas are taken over the cycle $ABCDA$, as shown in the $P-T$ diagram.The work done on the gas in taking it from $D$ to $A$ is

A
$+ 414$ $R$
B
$-690 $ $R$
C
$-690$ $ R$
D
$-414$ $ R$
(AIEEE-2009)
Solution
Work done by the system in the isothermal process
$DA\,is\,W = 2.303nRT\,{\log _{10}}\frac{{{P_D}}}{{{P_A}}}$
$ = 2.303 \times 2\,R \times 300\,{\log _{10}}\frac{{1 \times {{10}^5}}}{{2 \times {{10}^5}}} = – 414R$
Therefore work done on the gas is $+414\,R.$
Standard 11
Physics