Pressure of $1 \mathrm{~g}$ ideal gas $X$ at $300 \mathrm{~K}$ is $2 \mathrm{~atm}$. When $2 \mathrm{~g}$ of…
Pressure of $1 \mathrm{~g}$ ideal gas $X$ at $300 \mathrm{~K}$ is $2 \mathrm{~atm}$. When $2 \mathrm{~g}$ of another gas $Y$ is introduced in the same vessel at same temperature, the pressure become $\mathrm{l}$ atm. The correct relationship between molar masses of $X$ and $Y$ is
$M_Y=2 M_X$
$M_Y=4 M_X$
$M_x=4 M_y$
$M_X=5 M_Y$
Solution
Given,
Initially,
$\begin{array}{ll}p=2 \mathrm{~atm}, & T=300 \mathrm{~K} \\ V=v \mathrm{~L}, & m_x=1 \mathrm{~g}\end{array}$
Finally,
$\begin{array}{rlrl}p & =2+1 & T & =300 \mathrm{~K} \\ & =3 \mathrm{~atm} & m_y & =2 \mathrm{~g}\end{array}$
$\quad V=v \mathrm{~L}$
Let molar mass of gas $X$ and $Y$ be $M_x$ and $M_y$ respectively. Applying ideal gas equation to initial and final conditions.
Initially, $2 \times V=\frac{1}{M_x} R T$ ...(i)
Finally, $3 \times V=\left(\frac{1}{M_x}+\frac{2}{M_y}\right) R T$....(ii)
Dividing (i) by (ii), we get
$M_y=4 M_x$.