Question

A 1L sample of gas at 1atm pressure and 298K expands isothermally to 10L. It is then heated to 500K, compressed to 1L and then cooled 298K. What is the ∆U of overall the process?

Answer 1

∆U is zero for the isothermal process. Here we get ∆U value only for the seconded process.

From the first process we get the final pressure of the system

P₁/V₁ = P₂/V₂

P₂ = (1×1)/10

   = 0.1 atm

For contraction

W = P₂ × ∆V

   = 0.1 × 9 = 0.9 atm L = 91.1925 J

The process takes place at constant pressure P₂, hence, the heat absorbed by the system due to the change in the temperature

Q =  nc_p∆T

Here we consider a monatomic ideal gas, for which c_p = 5R/2.

  n = 1/22.4 = 0.045 mole, R=8.3145 J. K^-1. mol^-1

Q =  0.045× (5/2)×8.3145×202  J

   = 188.94 J

According to first law of thermodynamics

Q = -W + ∆U

∆U = Q + W = (188.94 + 91.19) J = 280.1 J (Answer)

∆ U of overall process  is 280.1 J.