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Grade upto college level Physical Chemistry

A sample of argon gas at 1 am pressure and 27oC expands reversibly and adiabatically from 1.25 dm3 to 2.50 dm3. Calculate the enthalpy change in this process.CV m for argon is 12.48 JK-1 mol-1.

Profile image of Shane Macguire
12 Years agoGrade upto college level
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1 Answer

Profile image of Deepak Patra
12 Years ago
Hello Student,
Please find the answer to your question
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
now we know that
∆H = nCP∆T = 0.05 * 20.8(188.5 - 300) = - 115.41 joules
[∵ CP = Cv + R = 12.48 + 3.314 = 20.8]

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jitender
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