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

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