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A thermodynamic system is taken form an initial state $i$ with internal energy $U_1=100 \ J$ to the final state along two different paths iaf and ibf, as schematically shown in the fire. The work done by the system along the paths $af$, ib and bf are $W _{ af }=200 \ J , W _{ ID }=50 \ J$ and $W _{ br }=100 \ J$ respectively. The heat supplied to the system along the path iaf, ib and bf are $Q_{\mid a t l} Q_{b r}$ and $Q_{10}$ respectively. If the internal energy of the sytem in the state $b$ is $U_b=$ $200 \ J$ and $Q_{l a t}=500 \ J$, the ratio $Q_{b J} / Q_{10}$ is:
$1$
$2$
$3$
$4$
Solution
$w _{\text {lot }}=150 \ J $
$w _{\text {lat }}=200 \ J $
$Q _{\text {lat }}=500 \ J So U _{\text {lat }}=300 \ J$
So : $U_t=400 \ J$
$U_b=100 \ J $
$Q_b=100+50=150 \ J $
$Q_{b t}=300+150=450 \ J$
So the required ratio $\frac{Q_{b f}}{Q_{i b}}=\frac{450-150}{150}=2$
