Question

A reversible compression of 1 mol of an ideal gas in a piston/cylinder device result in a pressure increase from 1 bar to P_2 and a temperature increase from 400 K to 950 K. The path followed by the gas during compression is given by PV^1.55 = const, and the molar heat capacity of the gas is given by C_p/R = 3.85+0.57 time sign 10^-3T [T/K] Determine the heat transferred during the process and the final pressure.

Answer 1

As,

PV=NRT

as no. of moles are constant

So

PV/T=Constant

(T/P) *constant=V

also

PV^1.55 = constant

P((T/P) *constant)^1.55=constant

P^(-0.55)*(T)^(1.55)= constant

P₂/P₁ = T₂/T₁

P_{2 =} T₂/T₁ * P1

AS, P₁ = 1

SO,

P_{2 =} 950/400

P2 = 2.375 Bars

Cp- Cv= R

C_P/R = 3.85 + 0.57 x10^-3 T

From first law of Thermodyanamics

the heat transferred is calculated as.

Q = W + dU

BY putting values In above equation,

We get the heat required for the process