JEE MAIN - Physics (2009 - No. 20)
Two moles of helium gas are taken over the cycle $$ABCD,$$ as shown in the $$P$$-$$T$$ diagram.
The work done on the gas in taking it from $$D$$ to $$A$$ is :
$$+414$$ $$R$$
$$-690$$ $$R$$
$$+690$$ $$R$$
$$-414$$ $$R$$
Explanation
Work done by the system in the isothermal process
$$DA$$ is $$W = 2.303nRT\,{\log _{10}}{{{P_D}} \over {{P_A}}}$$
$$ = 2.303 \times 2R \times 300{\log _{10}}{{1 \times {{10}^5}} \over {2 \times {{10}^5}}} = - 414R.$$
Therefore work done on the gas is $$ + \,414\,R.$$
$$DA$$ is $$W = 2.303nRT\,{\log _{10}}{{{P_D}} \over {{P_A}}}$$
$$ = 2.303 \times 2R \times 300{\log _{10}}{{1 \times {{10}^5}} \over {2 \times {{10}^5}}} = - 414R.$$
Therefore work done on the gas is $$ + \,414\,R.$$
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