Latest Articles

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
Work done by gas solved example

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
Work done by gas solved example

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

Also Read