Question

2 root 3 please give me answer​

Answer 1

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]

Step-by-step explanation: