Question

A perfect gas exerting a pressure P atm and has density r (gL^–1). A plot of (P r) Versus P at constant T is drawn if $\left\lbrack \frac{d}{dp}(\rho P) \right\rbrack_{p = 8.21atm}$= 5 then the value of T is [Molar mass of gas = 4 g mol^–1]

• A. 40 K
• B. 640 K
• C. 160 K
• D. 320 K

Answer 1

Answer: (C)

160 K

Answer 2

Q PM = rRT [M = ]

\ pr = P² $\left( \frac{M}{RT} \right)$

Now, $\frac{d(P\rho)}{dP}$ = $\frac{2PM}{RT}$ = 5

\ T = 160 K.