Which of the following gases of the same mass | vedclass

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

Question

(A) Let the mass of each gas be $m = 1 \ g$. The number of moles $n = \frac{m}{M}$,where $M$ is the molar mass. Since $m$ is constant,$n \propto \frac

Vedclass · Accepted Answer

$H_2$

Vedclass · Answer

(A) Let the mass of each gas be $m = 1 \ g$. The number of moles $n = \frac{m}{M}$,where $M$ is the molar mass. Since $m$ is constant,$n \propto \frac{1}{M}$. The molar masses are: $M(H_2) = 2 \ g/mol$,$M(N_2) = 28 \ g/mol$,$M(O_2) = 32 \ g/mol$,$M(Cl_2) = 71 \ g/mol$. Thus,the number of moles follows the order: $n(H_2) > n(N_2) > n(O_2) > n(Cl_2)$. From the ideal gas equation $PV = nRT$,at constant $T$ and $V$,$P \propto n$. Therefore,the gas with the highest number of moles will exert the highest pressure. Since $H_2$ has the lowest molar mass,it has the highest number of moles for a given mass,and thus exerts the highest pressure.

Which of the following gases of the same mass exerts the highest pressure at a constant temperature?

Similar Questions

Calculate the volume of gas at $1.25 \ atm$,if the volume occupied by the gas at $1 \ atm$ and at the same temperature is $25 \ mL$. (in $mL$)

An ideal gas at constant temperature and pressure expands,then its

What is the density of one mole of $He$ (molar mass $= 4 \ g \ mol^{-1}$) at $300 \ K$ and a pressure of $0.82 \ atm$? $(R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1})$

In a closed vessel,an ideal gas at $1 \, atm$ is heated from $27^{\circ} C$ to $327^{\circ} C$. The final pressure of the gas will approximately be $..... \, atm$

If the volume of a gas is $V \ L$ at $P \ atm$ pressure and $27 \ ^oC$ temperature,what will be the volume of the gas at $-173 \ ^oC$ temperature and $P \ atm$ pressure if the amount of gas is doubled?

Vedclass Test Series

Exam Paper Generator

Online Exam Module