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11.Thermodynamics
medium
The volume $( V)$ of a monatomic gas varies with its temperature $(T)$ , as shown in the graph. The ratio of work done by the gas , to the heat absorbed by it, when it undergoes a change from state $A$ to state $B$ , is
A
$\;\frac{2}{5}$
B
$\frac{2}{3}$
C
$\;\frac{2}{7}$
D
$\;\frac{1}{3}$
(NEET-2018)
Solution
Given process is isobaric.
$\therefore \,dQ = n{C_p}dT;\,where\,{C_p}\,is\,specific\,heat\,at\,constant\,pressure.$
or $dQ = n\left( {\frac{5}{2}R} \right)dT$
$Also,\,dW = PdV = nRdT\,\left( {PV = nRT} \right)$
Required ratio$ = \frac{{dW}}{{dQ}} = \frac{{nRdT}}{{n\left( {\frac{5}{2}R} \right)dT}} = \frac{2}{5}$
Standard 11
Physics
