Question

# One mole of an ideal monatomic gas is compressed irreversibly from 2.00 atmto 6.00 atmwhile being...

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.

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

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