A sample of argon gas at 1 am pressure and 27 o C expands reversibly and adiabatically from 1.25 dm 3 to 2.50 dm 3 . Calculate the enthalpy change in this process.C V m for argon is 12.48 JK -1 mol -1 .

Question

Deepak Patra · Answer

Hello Student,

Please find the answer to your question

For adiabatic expansion, we have

ℓn T₁/T₂ = R/C_v ℓn V₂/V₁

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

[∵ C_P = C_v + R = 12.48 + 3.314 = 20.8]

Thanks
jitender
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A sample of argon gas at 1 am pressure and 27 o C expands reversibly and adiabatically from 1.25 dm 3 to 2.50 dm 3 . Calculate the enthalpy change in this process.C V m for argon is 12.48 JK -1 mol -1 .

A sample of argon gas at 1 am pressure and 27^oC 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.

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A sample of argon gas at 1 am pressure and 27 o C expands reversibly and adiabatically from 1.25 dm 3 to 2.50 dm 3 . Calculate the enthalpy change in this process.C V m for argon is 12.48 JK -1 mol -1 .

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.

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A sample of argon gas at 1 am pressure and 27^oC 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.