In an adiabatic process, R=(2/3) Cv . The pressure of the gas will be proportional to:

Question

In an adiabatic process, R=(2/3) Cv . The pressure of the gas will be proportional to: (A) T5 / 3 (B) T5 / 2 (C) T5 / 4 (D) T5 / 6

Tardigrade Admin · Accepted Answer

R=(2/3) CV We know, Cp-CV=R or &gamma;-1=(R/Cv) or R=C v(&gamma;-1) Comparing &gamma;=(5/3) P1-&gamma; T&gamma;= constant = K or P propto T(&gamma; / &gamma;-1), Given &gamma;=(5/3) (&gamma;/(&gamma;-1))=(5/2) So, P propto T5 / 2

Q. In an adiabatic process, $R = \frac{2}{3} C_{v} .$ The pressure of the gas will be proportional to:

$T^{5/3}$

$T^{5/2}$

$T^{5/4}$

$T^{5/6}$

$R = \frac{2}{3} C_{V}$
We know, $C_{p} - C_{V} = R$
or $γ - 1 = \frac{R}{C _{v}}$ or $R = C v (γ - 1)$
Comparing $γ = \frac{5}{3}$
$P^{1 - γ} T^{γ} =$ constant $= K$
or $P \propto T^{(γ / γ - 1)}$ ,
Given $γ = \frac{5}{3}$
$\frac{γ}{( γ - 1 )} = \frac{5}{2}$
So, $P \propto T^{5/2}$

Q. In an adiabatic process, R=32​Cv​. The pressure of the gas will be proportional to:

T5/3

T5/2

T5/4

T5/6