Question

Two cylinders, both fitted with frictionless pistons, are filled with mixtures of He and Ar gases. In the first cylinder, the masses of He and Ar are \(m_1\) and \(m_2\), respectively. In the second cylinder, the masses of He and Ar are \(m_2\) and \(m_1\), respectively. The molar mass of Ar is \(10\) times the molar mass of He. The external pressure applied by the piston on the first cylinder needs to be \(5\) times that on the second cylinder so that the volume of the gas mixtures in both the cylinders are equal at the same temperature. Assuming He and Ar behave like ideal gases, the value of \(\left(\dfrac{m_1}{m_2}\right)\) is ____.

Hint:

Since volume and temperature are the same for both cylinders, use PV=nRT to turn the given pressure ratio directly into a mole ratio, n1/n2 = P1/P2. Write the total moles in each cylinder as the sum of moles of He and Ar, using the given relation between their molar masses. To save algebra, introduce the required ratio m1/m2 as a single unknown from the start rather than solving for m1 and m2 separately.

Answer 1

Concept: For ideal gases, \[ PV = nRT \] At constant temperature, \[ V \propto \frac{n}{P} \] Thus, if two gaseous systems have equal volume at the same temperature, \[ \frac{n_1}{P_1} = \frac{n_2}{P_2} \] The number of moles is calculated using: \[ n = \frac{\text{mass}}{\text{molar mass}} \] We are also given: \[ M_{\mathrm{Ar}} = 10M_{\mathrm{He}} \] This relation will help simplify the mole expressions.

Step 1: Calculating total moles in the first cylinder. Let molar mass of He be: \[ M \] Then molar mass of Ar is: \[ 10M \] In the first cylinder:
• Mass of He \(= m_1\)
• Mass of Ar \(= m_2\) Hence moles of He: \[ \frac{m_1}{M} \] and moles of Ar: \[ \frac{m_2}{10M} \] Therefore total moles in first cylinder are: \[ n_1 = \frac{m_1}{M} + \frac{m_2}{10M} \] Taking common denominator: \[ n_1 = \frac{10m_1 + m_2}{10M} \]

Step 2: Calculating total moles in the second cylinder. In the second cylinder:
• Mass of He \(= m_2\)
• Mass of Ar \(= m_1\) Thus moles of He: \[ \frac{m_2}{M} \] and moles of Ar: \[ \frac{m_1}{10M} \] Hence total moles are: \[ n_2 = \frac{m_2}{M} + \frac{m_1}{10M} \] \[ n_2 = \frac{10m_2 + m_1}{10M} \]

Step 3: Using the equal volume condition. The problem states: \[ P_1 = 5P_2 \] Also the volumes are equal and temperature is same. Using: \[ \frac{n_1}{P_1} = \frac{n_2}{P_2} \] Substitute \(P_1 = 5P_2\): \[ \frac{n_1}{5P_2} = \frac{n_2}{P_2} \] Cancel \(P_2\): \[ \frac{n_1}{5} = n_2 \] Thus, \[ n_1 = 5n_2 \]

Step 4: Substituting mole expressions. Using expressions for \(n_1\) and \(n_2\): \[ \frac{10m_1 + m_2}{10M} = 5\left( \frac{10m_2 + m_1}{10M} \right) \] Multiply both sides by \(10M\): \[ 10m_1 + m_2 = 5(10m_2 + m_1) \] Expand: \[ 10m_1 + m_2 = 50m_2 + 5m_1 \] Rearranging: \[ 10m_1 - 5m_1 = 50m_2 - m_2 \] \[ 5m_1 = 49m_2 \] \[ \frac{m_1}{m_2} = \frac{49}{5} \] But this contradicts the expected physical simplification. Let us carefully interpret the pressure statement again. The pressure on the first cylinder must be \(5\) times the second cylinder: \[ P_1 = 5P_2 \] For equal volume at same temperature: \[ \frac{n_1RT}{P_1} = \frac{n_2RT}{P_2} \] Thus, \[ \frac{n_1}{5P_2} = \frac{n_2}{P_2} \] \[ n_1 = 5n_2 \] Substituting correctly: \[ 10m_1 + m_2 = 5(10m_2 + m_1) \] \[ 10m_1 + m_2 = 50m_2 + 5m_1 \] \[ 5m_1 = 49m_2 \] Thus mathematically: \[ \boxed{\frac{m_1}{m_2} = \frac{49}{5}} \] Final Answer: \[ \boxed{\frac{49}{5}} \]

Answer 2

Concept:

• For an ideal gas, $PV=nRT$. At fixed volume $V$ and fixed temperature $T$, pressure is directly proportional to the total number of moles: $P \propto n$.
• Since the required quantity is itself a ratio, $m_1/m_2$, it is faster to introduce this ratio as a single unknown $x$ from the start, instead of tracking $m_1$ and $m_2$ separately.
• Any factor common to both mole expressions, such as the molar mass of He, cancels out of the ratio $n_1/n_2$ and never needs to be evaluated.

Step 1: Convert the pressure condition into a mole ratio.
Both cylinders have equal volume and equal temperature, so from $PV=nRT$, $P \propto n$ for each cylinder.
Given $P_1=5P_2$, this directly gives $\dfrac{n_1}{n_2}=5$.

Step 2: Write the mole expressions and introduce $x=\dfrac{m_1}{m_2}$.
Let the molar mass of He be $M$, so the molar mass of Ar is $10M$ (given).
Cylinder 1 has mass $m_1$ of He and $m_2$ of Ar: $n_1=\dfrac{m_1}{M}+\dfrac{m_2}{10M}$
Cylinder 2 has mass $m_2$ of He and $m_1$ of Ar: $n_2=\dfrac{m_2}{M}+\dfrac{m_1}{10M}$

Step 3: Divide both expressions by the common factor $\dfrac{m_2}{M}$.
Dividing $n_1$ and $n_2$ by the same nonzero quantity leaves their ratio unchanged, and cancels $M$ immediately:
$\dfrac{n_1}{m_2/M}=\dfrac{m_1}{m_2}+\dfrac{1}{10}=x+\dfrac{1}{10}$
$\dfrac{n_2}{m_2/M}=1+\dfrac{m_1}{10m_2}=1+\dfrac{x}{10}$
So: $\dfrac{n_1}{n_2}=\dfrac{x+\frac{1}{10}}{1+\frac{x}{10}}$

Step 4: Solve for $x$ using $\dfrac{n_1}{n_2}=5$.
$\dfrac{x+0.1}{1+\frac{x}{10}}=5$
$x+0.1=5\left(1+\dfrac{x}{10}\right)=5+\dfrac{x}{2}$
$x-\dfrac{x}{2}=5-0.1$
$\dfrac{x}{2}=4.9$
$x=9.8=\dfrac{49}{5}$

Final Answer: $\dfrac{m_1}{m_2}=\dfrac{49}{5}=9.8$