Question

Calculate the values of q, w, delta U, and delta H for the reversible adiabatic expansion of 1 mole of a monoatomic ideal gas from 5.00 m^3 to 25.0 m^3. The temperature of the gas is initially 298K.

Answer 1

for adiabatic process q = 0

we need to find the final temperature of the process with the next equation

T₂ / T₁ = (V₁ / V₂) ^{- 1}

for monoatomi gas is 5 / 3 so - 1 = 2 / 3

T2 / 298 = (5 / 25 ) ^2/3

T2 / 298 = 0.34199

T2 = 0.34199 * 298 = 101.9 K or 102 K

For

since Q = 0

= n Cv *

Cv is 3/2 R for monoatomic gases , cv is heat capacity at constant volume

W = 1 * 3 / 2 * 8.314 J / K mol * (102 - 298) =- 2456 J or - 2.45 KJ

For Enthalpy apply

Cp = 5/2 R; for monoatomic gases, cp is molar heat capacity at constant pressure

Enthalpy = 1 * 5/2 * 8.314 * (101 - 298) = -4094.645 J or -4.09 KJ