# Work done by gas

## Work done by gas for Volume changes

• Consider a cylinder filled with gas and equipped with a movable piston as shown in fig below

fig - Force exerted by a system during small expansion.
Suppose,
A - Cross Sectional area of cylinder
P - Pressure exerted by piston at the piston face.
PA - Force exerted by the system.
• If piston moves out by a distance dx then Work done by this force is dW given by
$dW = PAdx$
$= PdV$ ---(1)
since $V = Adx$ and dV is change in volume of the system.
• In a finite volume change from V1 to V2
$W=\init PdV$ -----(2)
where limits of integration goes from V1 to V2
Graphically this relationship is shown below

• Thus eqn (2) can be interpreted graphically as area under the curve between limits V1and V2.
• If pressure remains constant while the volume changes, then work is
W = P(V2-V1)(3)
• Work done not only depends on initial and final states but also on the intermediate states i.e., on the path.
Learning:Work done in a process is given by area under the process on the PV diagram

## Solved Example

Question 1
A ideal monatomic gas is taken through the process ABCDA as shown in below P-V diagram. Calculate the Work done by gas

Solution
Work done by the gas from A to B = 0 ( as volume constant and area enclosed by AB is zero)
Work done by the gas from B to C = 2PV ( Area enclosed by line BC)
Work done by the gas from C to D = 0 ( as volume constant and area enclosed by AB is zero)
Work done by the gas from D to A = -PV ( Area enclosed by line BC and negative as volume is decreased)
Net Work Done = 2PV - PV = PV

Question 2
An ideal gas undergoes cyclic process ABCDA as shown in given P-V diagram .The amount of work done by the gas is

Solution
Work done by the gas from A to B = 2PV ( Area enclosed by line AB)
Work done by the gas from B to C = 0 ( as volume constant and area enclosed by BC is zero)
Work done by the gas from C to D = -4PV ( Area enclosed by line CD and negative as volume is decreased)
Work done by the gas from D to A = 0 ( as volume constant and area enclosed by DA is zero)
Net Work Done = 2PV - 4PV = -2PV

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