Question
Derive the expression for PV work.
✓
Answer
Consider a certain amount of an ideal gas enclosed in an ideal cylinder fitted with massless, frictionless rigid movable piston at pressure P, occupying volume $V_1$ at temperature T.
As the gas expands, it pushes the piston upward through a distance $d$ against external force $F$, pushing the surroundings.
The work done by the gas is,
$W =$ opposing force $\times$ distance $=- F \times d$
-ve sign indicates the lowering of energy of the system during expansion.
If $a$ is the cross section area of the cylinder or piston, then,
$
W =-\frac{F}{a} \times d \times a
$
Now the pressure is $P _{ ex }=\frac{F}{a}$
while volume change is, $\Delta V=d \times a$
$\therefore W =- P _{ ex } \times \Delta V$
If during the expansion, the volume changes from $V_1$ and $V_2$ then, $\Delta V=$ $V _2- V _1$
$\therefore W =- P _{ ex }\left( V _2- V _1\right)$
During compression, the work W is +ve, since the energy of the system is increased,
$
W =+ P _{ ex }\left( V _2- V _1\right)
$
As the gas expands, it pushes the piston upward through a distance $d$ against external force $F$, pushing the surroundings.
The work done by the gas is,
$W =$ opposing force $\times$ distance $=- F \times d$
-ve sign indicates the lowering of energy of the system during expansion.
If $a$ is the cross section area of the cylinder or piston, then,
$
W =-\frac{F}{a} \times d \times a
$
Now the pressure is $P _{ ex }=\frac{F}{a}$
while volume change is, $\Delta V=d \times a$
$\therefore W =- P _{ ex } \times \Delta V$
If during the expansion, the volume changes from $V_1$ and $V_2$ then, $\Delta V=$ $V _2- V _1$
$\therefore W =- P _{ ex }\left( V _2- V _1\right)$
During compression, the work W is +ve, since the energy of the system is increased,
$
W =+ P _{ ex }\left( V _2- V _1\right)
$
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