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  4. An ideal monoatomic gas undergoes a process A to B as shown in the figure. <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/physics/e0fd8b665b4380a654bd3badc29bcb4a-.png /> If the heat supplied to the gas is 18P0V0, then the shaded area is equal to

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Q. An ideal monoatomic gas undergoes a process A to B as shown in the figure.
image If the heat supplied to the gas is $18P_0V_0$, then the shaded area is equal to

Solution:

$\Delta Q = \Delta U +W$
$W = area \,under\,PV \, curve = \Delta Q - \Delta U$
$= 18P_0V_0-nC_v\Delta T$
$= 18P_0V_0-\frac {3}{2}nR\Delta T$
$W = 18P_0V_0-\frac {3}{2} (P_2V_2-P_1V_1)$ $= 18P_0V_0-\frac {3}{2} (9P_0V_0-2P_0V_0)$
$= 18 P_0V_0 - \frac {21}{2}P_0V_0$
$= 75 P_0V_0$