One mole of a diatomic gas undergoes a thermodynamic process, whose process equation is P propto V2 . The molar specific heat of the gas is

Question

One mole of a diatomic gas undergoes a thermodynamic process, whose process equation is P propto V2 . The molar specific heat of the gas is (A) (17 R/3) (B) (17 R/6) (C) (15 R/4) (D) (15 R/8)

Tardigrade Admin · Accepted Answer

Given PV- 2= constant Comparing with PVN= constant, we get ∴ N=-2 C=Cv+(R/1 - N)=(5 R/2)+(R/1 + 2)=(5 R/2)+(R/3) =(15 R + 2 R/6)=(17 R/6)

Q. One mole of a diatomic gas undergoes a thermodynamic process, whose process equation is $P \propto V^{2}$ . The molar specific heat of the gas is

$\frac{17 R}{3}$

$\frac{17 R}{6}$

$\frac{15 R}{4}$

$\frac{15 R}{8}$

Given $P V^{- 2} =$ constant
Comparing with $P V^{N} =$ constant, we get
$∴ N = - 2$
$C = C_{v} + \frac{R}{1 - N} = \frac{5 R}{2} + \frac{R}{1 + 2} = \frac{5 R}{2} + \frac{R}{3}$
$= \frac{15 R + 2 R}{6} = \frac{17 R}{6}$

Q. One mole of a diatomic gas undergoes a thermodynamic process, whose process equation is P∝V2 . The molar specific heat of the gas is

317R​

617R​

415R​

815R​