Question

An ideal monoatomic gas undergoes a process A to B as shown in the figure.
If the heat supplied to the gas is $18P_0{V}_0$, then the shaded area is equal to

• A. $7.5P_0{V}_0$
• B. $5.5P_0{V}_0$
• C. $10.5P_0{V}_0$
• D. $P_0{V}_0$

Answer 1

Answer: (A)

$\Delta Q=\Delta U+W$
$W=areaunderPVcurve=\Delta Q-\Delta U$
$=18P_0{V}_0-n{C}_v\Delta T$
$=18P_0{V}_0-\frac{3}{2}nR\Delta T$
$W=18P_0{V}_0-\frac{3}{2}(P_2{V}_2-P_1{V}_1)$ $=18P_0{V}_0-\frac{3}{2}(9P_0{V}_0-2P_0{V}_0)$
$=18P_0{V}_0-\frac{21}{2}{P}_0{V}_0$
$=75P_0{V}_0$