A gas occupies 2 liter at STP. It provides 300 joule heat so that its volume becomes 2.5 liters at 1 atm. Calculate change in the internal energy.

Work done = -P × dV                               .'. Work is carried out at constant P and thus irreversible =−[1×(2.5−2.0)]                = - 0.5 liter atm                =−0.5×1.987×4.1840.0821 joule =−50.63JFrom first law of thermodynamics∴ q=ΔU−W300=ΔU+50.63ΔU=249.37 joule.