Question

The pressure of $1g$ of an ideal gas $A$ at $300K$ is found to be $2bar$. When $2g$ of another ideal gas $B$ is introduced in the same flask at the same temperature, the pressure becomes $3bar$. Which of the following is correct?
($m_A=$mass of gas $A$; $m_B$ mass of gas $B$)

• A. $m_B=4m_A$
• B. $m_A=3m_B$
• C. $m_B=3m_A$
• D. $m_A=m_B$

Answer 1

The correct option is A $m_B=4m_A$

Given,
$P_A=\frac{n_{A}RT}{V}=\frac{w_{A}RT}{m_{A}V}=\frac{1\times RT}{m_{A}V}=2bar$
$P_A+P_B=$ 3 bar
$\therefore P_B=3-2=1$ bar
$P_B=\frac{w_{B}RT}{m_{B}V}=\frac{2RT}{m_{B}V}=1\Rightarrow \frac{P_B}{P_A}=\frac{2m_A}{m_B}=\frac{1}{2}$
Thus, $m_B=4m_A$