Question

If a gas is compressed from 10.0L to 3.0L at constant temperature and 2 atm pressure and releases 56.5 kJ of heat to the surroundings at the same time, what is ΔE for the gas?

Answer 1

Work done , W = -P dV

Where P = Pressure = 2 atm x(101325 pas/ atm )

              = 202650 Pas

         dV = (3.0 - 10.0)L = -7.0 L x (10^-3 m³/L)

             = -7.0x10^-3 m³

So W = 1418.6 J

         = 1.42 kJ

Amount releases to the surroundings is 56.5 kJ

So Q = -56.5 kJ

From first law of thermodynamics ,

ΔE = Q + W

     = -56.5 kJ + 1.42 kJ

     = -55.08 kJ