Class 12 Maharashtra State Board

A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm from a volume of 5 litres to a volume of 10 litres. In doing so it absorbs 400 J of thermal energy from its surroundings. Determine the change in internal energy of system.

As per the given question the given data can be listed as,

Pressure P = 1 atm = 1.013 x 10⁵ Pa

$https://latex.codecogs.com/svg.image?Initial volume V_{1}= 5 litre = 5 x 10^{-3}m^{3}$

$https://latex.codecogs.com/svg.image?Final volume V_{2}= 10 litre = 10 x 10^{-3}m^{3}$

Q = 400J

To find, change in internal energy dU =?

Solution -

Work done by the system on its surrounding is,

$https://latex.codecogs.com/svg.image?W = P(V_{2}-V_{1})$

$https://latex.codecogs.com/svg.image?W =1.013 \mathrm{x} 10^{5}(10 \mathrm{x 10^{-3}-5\mathrm{x}10^{-3}})$

$https://latex.codecogs.com/svg.image?W =1.013 \mathrm{x} 10^{5}(5 \mathrm{x 10^{-3})$

W = 1.013 x 500 = 506.5 J

From first of thermodynamics

we can write the change in internal energy of the system as

dU = Q - W

dU = 400J - 506.5J = -106.5 J

therefore we can say that there is a decrease in the internal energy by 106.5J

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