Question

One mole of an ideal monoatomic gas is taken round cyclic process $ABCA$ as shown in figure. Calculate the heat rejected by the gas in the path $CA$ the heat absorbed by the gas in the path $AB$ respectively.

• A. $2.5P_o{V}_o,3P_o{V}_o$
• B. $3P_o{V}_o,2.5P_o{V}_o$
• C. $5P_o{V}_o,5P_o{V}_o$
• D. $3P_o{V}_o,5P_o{V}_o$

Answer 1

The correct option is C $5P_o{V}_o,5P_o{V}_o$

The First Law of Thermodynamics states that energy can be converted from one form to another with the interaction of heat, work and internal energy, but it cannot be created nor destroyed, under any circumstances. Mathematically, this is represented as

$\Delta H=\Delta U+W$

with,

$\Delta H$ is the total change in internal energy of a system,

$\Delta U$ is the heat exchanged between a system and its surroundings, and

$W$ is the work done by or on the system.

For a monoatomic gas,

$\Delta U=\frac{3}{2}RT\left(Foronemole\right)$

Along CA, Temperature is not constant,

$PV=RTP=P_{o}RT=P_{o}VdU=\frac{3}{2}{P}_{o}dVIntegrate,weget\Delta U=\frac{3}{2}{P}_o[3V_o-V_o]=3{P_{o}V}_o$

$W=P_o[3V_o-V_o]=2{P_{o}V}_o$

$Q=\Delta U+W=5{P_{o}V}_o$

Similarly Along AB,

$\Delta U=\frac{3}{2}{P}_o[3V_o-V_o]=3{P_{o}V}_o$

$W=P_o[3V_o-V_o]=2{P_{o}V}_{o}Q=\Delta U+W=5{P_{o}V}_o$