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11.Thermodynamics
hard
$1 \,\,kg$ of a gas does $20\,\, kJ$ of work and receives $16 \,\,kJ$ of heat when it is expanded between two states. $A$ second kind of expansion can be found between the initial and final state which requires a heat input of $9\,\, kJ$. The work done by the gas in the second expansion is ....... $kJ$
A
$32$
B
$5$
C
$- 4$
D
$13$
Solution
For first kind of expansion
$\Delta W=20 \mathrm{KJ}$
$\Delta Q=16 \mathrm{KJ}$
$\therefore \Delta U=\Delta Q-\Delta W=-4 K J$
since, $U$ is a state function. Therefore, value of $DU$ in both expansions remain same.
Thus, for second expansion $:-$
$\Delta U=-4 K J, \Delta Q=9 K J$
$\therefore$ By first law of the thermodynamics
$\Delta \mathrm{W}=\Delta \mathrm{Q}-\Delta \mathrm{U}=13 \mathrm{KJ}$
Standard 11
Physics