Question

A sample of argon gas at 1 atm pressure and 27oC expands reversibly and adiabatically from 1.25 dm3 to 2.50 dm3. (Cv for argon gas is 16.5 j/k mol) the enthalpy change for the given system is

Answer 1

Answer:

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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Explanation:

Answer 2

hey

change in heat (q) = 0 or q remain constant. (iii) .... A gas expands by 0.5 litre against a constant pressure one ..... reversibly and adiabatically from 20 dm 3 to 10