Question

One mole of an ideal monoatomic gas is taken round cyclic process $ABCA$ as shown in figure. Calculate the work done by the gas.

• A. $P_o{V}_o$
• B. $-P_o{V}_o$
• C. $2P_o{V}_o$
• D. $-2P_o{V}_o$

Answer 1

The correct option is B $-P_o{V}_o$

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

$\Delta H=\Delta U+\Delta nRT$

with,

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

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

$\Delta nRT$ is the work done by or on the system.

For a cyclic process, change in internal energy is zero. So, the work done during the process is equal to the area under the graph. The area under the graph is equal to the area of the triangle.

$Area=\frac{1}{2}\times \left(Base\right)\times \left(Height\right)$

$=\frac{1}{2}\times (3P_o-P_o)\times (3V_o-V_o)=2P_o{V}_o$

In the cyclic process, work done is equal to the area of a closed curve. It is positive if the cycle is clockwise and it is negative if the cycle is anticlockwise.

$Work=2P_o{V}_o$

Hence, the correct option is $A$