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An ideal gas is taken through a process . The ratio of heat absorbed during AB to the work done on the gas during BC is ---

$(a)\;\large\frac{15}{16 ln 2}\qquad(b)\;\large\frac{16}{15 ln 2}\qquad(c)\;\large\frac{15}{16 }\qquad(d)\;\large\frac{16}{15}$

1 Answer

Answer : $\;\large\frac{15}{16 ln 2}$
Explanation :
Heat absrobed during the process $\;AB=n C_{p} \bigtriangleup T\quad$ (p is constant during AB )
$Q=n C_{p}\;(\large\frac{3T_{0}}{2})$
Work done on the gas during BC
$= nR (2T_{0}) ln (4)$
$w=4nRT_{0} ln(2)$
$\large\frac{d}{w}= \large\frac{nC_{p} (\large\frac{3T_{0}}{2})}{4n RT_{0} ln(2)}$
$=\large\frac{3}{8} \; . \large\frac{5}{2 ln2}=\large\frac{15}{16 ln 2}$
answered Mar 13, 2014 by yamini.v

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