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Question:

A gas is allowed to expand in a well insulated container against constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be:

+505 J

+1136.25 J

− J

− J

Solution:

The change in the internal energy is given by the following expression.
ΔU = q + W
Since container is well insulated, q = 0.
Hence, ΔU = W = −PΔV
Here, P = 2.5 atm, ΔV = 4.50 L − 2.50 L = 2.0 L
W = −(2.5 atm)(2.0 L) = −5.0 L atm
Since 1 L atm = 101.325 J,
W = −5.0 L atm × 101.325 J/L atm = −506.625 J ≈ −505 J
Therefore, ΔU = −505 J
Since the internal energy change is negative, work is done by the system. The correct answer is +505 J