Question

A sample of argon gas at 1 atm pressure and 27°C expands reversibly and adiabatically from 1.25 dm³ to 2.50 dm³. Calculate the enthalpy change in this process. $C_{V, m}$ for argon is 12.48 JK⁻¹ mol⁻¹.

Answer 1

Answer:

Explanation

Pressure = 1 atm

Temperature = 27°C

Expansion = 1.25 dm³ to 2.50 dm³

For adiabatic expansion, we have

ℓn T1/T2 = R/Cv ℓn V2/V1

and ∆H = nCP ∆T.

ℓn 300/T2 = 8.31/12.48 ℓn 2.50/1.25

solving, we get, T2 = 188.5 K

No. of moles of argon gas, N = PV/RT = 1 *1.25/0.082 *300 = 0.05

Thus,

∆H = nCP∆T = 0.05 * 20.8(188.5 - 300) = - 115.41 joules

[∵ CP = Cv + R = 12.48 + 3.314 = 20.8]