Question

One mole of a monoatomic ideal gas is taken along ABCA as shown in the diagram. The net heat absorbed by the gas in the given cycle is

• A. $PV$
• B. $\frac{PV}{2}$
• C. $2PV$
• D. $4PV$

Answer 1

The correct option is B $\frac{PV}{2}$

Work done by gas = Area under PV graph for the process.
$w=\int PdV$
For the given graph
$w=\frac{1}{2}(2P-P)(2V-V)=\frac{1}{2}PV$
The first law of thermodynamics gives us the relation:
$q=\Delta U+w$
For a cyclic process, $\Delta U=0$
Hence q = w = $\frac{1}{2}PV$
Hence option B is correct.