A container of volume 2.24 , L can withstand | vedclass

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

Question

(D) Using the ideal gas equation $pV = nRT$,where $p = 2 \, atm$,$V = 2.24 \, L$,$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,and $T = 298

Vedclass · Accepted Answer

$4.2$

Vedclass · Answer

(D) Using the ideal gas equation $pV = nRT$,where $p = 2 \, atm$,$V = 2.24 \, L$,$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,and $T = 298 \, K$.$n = \frac{pV}{RT} = \frac{2 	imes 2.24}{0.0821 	imes 298} \approx 0.183 \, mol$.The molar mass of $N_2$ is $28 \, g/mol$.The maximum mass of $N_2$ is $m = n 	imes M = 0.183 	imes 28 \approx 5.124 \, g$.Since the container explodes at $2 \, atm$,the safe amount must be slightly less than $5.124 \, g$. Among the given options,$4.2 \, g$ is the closest value that ensures safety.

$A$ container of volume $2.24 \, L$ can withstand a maximum pressure of $2 \, atm$ at $298 \, K$ before exploding. The maximum amount of nitrogen (in $g$) that can be safely put in this container at this temperature is closest to

Similar Questions

Pressure of $1 \ g$ ideal gas $X$ at $300 \ K$ is $2 \ atm$. When $2 \ g$ of another gas $Y$ is introduced in the same vessel at the same temperature,the total pressure becomes $3 \ atm$. The correct relationship between molar masses of $X$ and $Y$ is:

State Avogadro's Law and Gay-Lussac's Law.

Assuming the molar mass of air is $28.84 \, g \, mol^{-1}$,calculate the payload (lifting capacity) of a balloon of volume $10^6 \, L$ filled with $He$ at $27 \, ^\circ C$ and $1 \, atm$ pressure.

If $P$,$V$,$M$,$T$ and $R$ are pressure,volume,molar mass,temperature and gas constant respectively,then for an ideal gas,the density is given by

$2 \ g$ of $O_2$ at $27 \ ^\circ C$ and $760 \ mm$ of $Hg$ pressure has volume ............. $L$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module