- Home
- Standard 11
- Physics
One mole of a gas obeying the equation of state $P(V-b)=R T$ is made to expand from a state with coordinates $\left(P_{1}, V_{1}\right)$ to a state with $\left(P_{2}, V_{2}\right)$ along a process that is depicted by a straight line on a $P-V$ diagram. Then, the work done is given by
$\frac{1}{2}$(${P_1} + {P_2})\left( {{V_2} - {V_1}} \right)$
$\;\frac{1}{2}$(${P_2} - {P_1})\left( {{V_2} - {V_1}} \right)$
$\frac{1}{2}$(${P_1} + {P_2})\left( {{V_2} - {V_1} + 2b} \right)$
$\;\frac{1}{2}$(${P_2} - {P_1})\left( {{V_2} + {V_1} + 2b} \right)$
Solution
Workdone during the complete cycle is equal to the area enclosed by the P-V graph
$W=\frac{1}{2} \text { base } \times \text { height }+ \text { Area of rectangular }$
$=\frac{1}{2}\left(V_2-V_1\right) \times\left(P_1-P_2\right)+\left(V_2-V_1\right) P_2$
$=\left(V_2-V_1\right)\left[\frac{P_1}{2}-\frac{P_2}{2}+P_2\right]$
$=\left(V_2-V_1\right)\left[\frac{P_1}{2}+\frac{P_2}{2}\right]$
$=\frac{1}{2}\left(P_1+P_2\right)\left(V_2-V_1\right)$
