Gujarati
Hindi
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

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