Question

One mole of an ideal monatomic gas is compressed irreversibly from 2.00 atmto 6.00 atmwhile being cooled from 400. K to 300. K. Calculate ?U, ?H, and ?S for this process.

Answer 1

For an ideal gas

At step 1

For isothermal process

Change in internal energy H = U = 0

Entropy change

= 1 mol x 8.314 J/mol·K x ln (2/6)

= - 9.130 J/K

At step 2

Change in enthalpy

= (5/2) x 1 mol x 8.314 J/mol·K x (300 - 400)K

= - 2.08 x 10^3 J

Change in internal energy

= - 2.08 x 10^3 J - 1 mol x 8.314 J/mol·K x (300 - 400)K

= - 1.25 x 10^3 J

Change in entropy

= (5/2) x 1 mol x 8.314 J/mol·K x ln ( 300/400)

= - 5.98 J/K

For combined process

Change in enthalpy

H = H for step1 - H for step 2

= 0 - 2.08*10^3 J

= 2.08 x 10^3 J

Change in internal energy

U = U for step1 - H for step 2

= 0 - 1.25 x 10^3

= - 1.25 x 10^3 J

Change in entropy

S = S for step1 - S for step 2

= - 9.130 - 5.98

= - 15.1 J/K