A monoatomic gas undergoes a process in which the pressure (.P.) and the volume (.V.) of the gas are related as PV- 3= constant. What will be the molar heat capacity of gas for this process?

Question

A monoatomic gas undergoes a process in which the pressure (.P.) and the volume (.V.) of the gas are related as PV- 3= constant. What will be the molar heat capacity of gas for this process? (A) (5 R/3) (B) (3/5)R (C) (7/4)R (D) (R/4)

Tardigrade Admin · Accepted Answer

PV- 3=constant n=-3 C=(R/&gamma; - 1)-(R/n - 1) &gamma; =(5/3) C=(R/(5)3 - 1)-(R/- 3 - 1) =(3 R/2)+(R/4) =(7/4)R

Q. A monoatomic gas undergoes a process in which the pressure $(P)$ and the volume $(V)$ of the gas are related as $P V^{- 3} =$ constant. What will be the molar heat capacity of gas for this process?

$\frac{5 R}{3}$

$\frac{3}{5} R$

$\frac{7}{4} R$

$\frac{R}{4}$

$P V^{- 3} = co n s t an t n = - 3$
$C = \frac{R}{γ - 1} - \frac{R}{n - 1} γ = \frac{5}{3}$
$C = \frac{R}{\frac{5}{3} - 1} - \frac{R}{- 3 - 1}$
$= \frac{3 R}{2} + \frac{R}{4}$
$= \frac{7}{4} R$

Q. A monoatomic gas undergoes a process in which the pressure (P) and the volume (V) of the gas are related as PV−3= constant. What will be the molar heat capacity of gas for this process?

35R​

53​R

47​R

4R​

PV−3=constantn=−3 C=γ−1R​−n−1R​γ=35​ C=35​−1R​−−3−1R​ =23R​+4R​ =47​R