Find the pressure of neon gas having density | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The ideal gas equation is given by $PV = nRT$. Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M} RT$. Rearranging for pressure $P$,we get $P = \fr

Vedclass · Accepted Answer

$1.309$

Vedclass · Answer

(B) The ideal gas equation is given by $PV = nRT$. Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M} RT$. Rearranging for pressure $P$,we get $P = \frac{m}{V} 	imes \frac{RT}{M} = \frac{dRT}{M}$. Here,density $d = 0.9 \ g \ L^{-1}$,$R = 8.34 	imes 10^{-2} \ bar \ L \ K^{-1} \ mol^{-1}$,$T = 350 \ K$,and the molar mass of neon $M = 20.18 \ g \ mol^{-1}$. Substituting the values: $P = \frac{0.9 	imes 8.34 	imes 10^{-2} 	imes 350}{20.18} \approx 1.309 \ bar$.

Find the pressure of neon gas having density $0.9 \ g \ L^{-1}$ at $350 \ K$ temperature. $(R = 8.34 \times 10^{-2} \ bar \ L \ K^{-1} \ mol^{-1})$ (in $bar$)

Similar Questions

The nitrogen gas pressure inside a container of volume $2.6 \ cm^3$ is $2.3 \ atm$ at $27^{\circ} C$. The approximate number of moles of nitrogen present in the container is $[R=0.0821 \ L \ atm \ mol^{-1} \ K^{-1}]$

The compressibility factor of a gas is defined as $Z = PV/RT$. The compressibility factor of an ideal gas is:

At $300 \, K$ temperature and $1 \, atm$ pressure,the volume of $7.0 \, g$ of a gas is $4.1 \, L$. What is the molar mass of the gas?

If the density of dioxygen gas at $STP$ is $1.43 \, g \, L^{-1}$,calculate its density at $17 \, ^\circ C$ and $800 \, torr$ pressure.

Under what conditions does an ideal gas have maximum density?

Vedclass Test Series

Exam Paper Generator

Online Exam Module