Two identical containers A and B with frictionless pistons contain the same ideal gas at the same te

Q: Why did we take the magnitude of the pressure change?

Since the volume increases during expansion, the final pressure is less than the initial pressure. The problem states the 'changes in pressure' as positive values ($\Delta p$ and $1.5 \Delta p$), which implies we should consider the magnitude of the decrease.

Q: Why is the temperature constant?

The problem explicitly states that the gas is allowed to expand 'isothermally'. By definition, an isothermal process occurs at a constant temperature.

Q: How do we know the molar mass is the same for both?

The problem mentions that both containers contain the 'same ideal gas'. Therefore, their molar mass $M$ must be identical.

Q: Could we have used the formula $p_1 V_1 = p_2 V_2$?

Yes, absolutely! For an isothermal process, $p_i V = p_f (2V)$, which gives $p_f = p_i / 2$. The change in pressure is $p_i - p_f = p_i / 2$. Since $p_i = nRT/V$, the change is $nRT/2V$.

Q: What if the gases were different?

If the gases were different, say Gas 1 and Gas 2, their molar masses $M_A$ and $M_B$ would be different. The ratio of moles $n_A/n_B$ would then be $(m_A/M_A) / (m_B/M_B)$, and we wouldn't be able to find a direct relation between $m_A$ and $m_B$ without knowing their molar masses.

Question

Two identical containers $A$ and $B$ with frictionless pistons contain the same ideal gas at the same temperature and the same volume $V$. The mass of the gas in $A$ is $m_A$ and that in $B$ is $m_B$. The gas in each cylinder is now allowed to expand isothermally to the same final volume $2V$. The changes in the pressure in $A$ and $B$ are found to be $\Delta p$ and $1.5 \Delta p$ respectively. Then

Options:

(A) $4 m_A = 9 m_B$

(B) $2 m_A = 3 m_B$

(C) $3 m_A = 2 m_B$

(D) $9 m_A = 4 m_B$

Sigma Percentile · Accepted Answer

Welcome to this fascinating problem from Thermodynamics! At first glance, it might seem like a lot of variables are thrown at us, but let's break it down step-by-step. Imagine you are in a lab with two identical cylinders, A and B. Both are fitted with frictionless pistons, meaning no energy is lost to heat from rubbing.

Inside these cylinders, we have the exact same ideal gas. They start at the exact same temperature $T$ and the exact same volume $V$. The only difference? The amount of gas. Cylinder A has a mass $m_A$, and cylinder B has a mass $m_B$.

### Analyzing the Setup

Now, we let the gases expand. But not just any expansion—an **isothermal** expansion. This is a crucial keyword! Isothermal means the temperature remains absolutely constant throughout the process. The pistons move up until the volume in both cylinders doubles to $2V$.

Because the volume increases while the temperature stays the same, the pressure must drop. The problem tells us that the magnitude of this pressure drop is $\Delta p$ for cylinder A and $1.5 \Delta p$ for cylinder B. Our goal is to find the relationship between their masses.

### The Master Equation

To connect pressure, volume, temperature, and mass, we need our trusty tool: the Ideal Gas Law.

$$p = \frac{nRT}{V}$$

Let's apply this to cylinder A. The initial pressure is $p_{iA} = \frac{n_A RT}{V}$. After expanding to $2V$, the final pressure is $p_{fA} = \frac{n_A RT}{2V}$.

The change in pressure (magnitude) is the difference between the initial and final states:

$$|\Delta p_A| = p_{iA} - p_{fA} = \frac{n_A RT}{V} - \frac{n_A RT}{2V} = \frac{n_A RT}{2V}$$

We are given that this change is equal to $\Delta p$.

Now, let's do the exact same thing for cylinder B. The initial pressure is $p_{iB} = \frac{n_B RT}{V}$ and the final pressure is $p_{fB} = \frac{n_B RT}{2V}$.

The change in pressure here is:

$$|\Delta p_B| = \frac{n_B RT}{2V}$$

And we are told this change is equal to $1.5 \Delta p$.

### Final Calculation

We have two beautiful equations now. Let's take their ratio to eliminate all the common terms.

$$\frac{|\Delta p_A|}{|\Delta p_B|} = \frac{\frac{n_A RT}{2V}}{\frac{n_B RT}{2V}}$$

Notice how the $R$, $T$, and $2V$ terms completely cancel out! We are left with:

$$\frac{\Delta p}{1.5 \Delta p} = \frac{n_A}{n_B}$$

$$\frac{1}{1.5} = \frac{n_A}{n_B} \Rightarrow \frac{n_A}{n_B} = \frac{2}{3}$$

We know that the number of moles $n$ is simply the mass $m$ divided by the molar mass $M$. Since both cylinders contain the **same gas**, their molar mass $M$ is identical.

$$\frac{m_A / M}{m_B / M} = \frac{2}{3}$$

$$\frac{m_A}{m_B} = \frac{2}{3}$$

Cross-multiplying gives us our final, elegant result:

$$3m_A = 2m_B$$

And there you have it! By carefully applying the ideal gas law and understanding the nature of an isothermal process, we've unraveled the relationship between the masses. Keep practicing, and these concepts will become second nature to you!

Select Answer:

Visualized Solution

\text{Initial Setup}

\text{Isothermal Expansion}

\text{Pressure Change in A}

\text{Pressure Change in B}

\text{Ratio of Pressure Changes}

\text{Mass Relationship}

\text{Conclusion}

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

An ideal gas expands isothermally from a volume $V_{1}$ to $V_{2}$ and then compressed to original volume $V_{1}$ adiabatically. Initial pressure is $p_{1}$ and final pressure is $p_{3}$ . The total work done is $W$ . Then,

An ideal gas is expanding such that $p T^{3} = constant$ . The coefficient of volume expansion of the gas is

An ideal gas is expanding such that $p T^{2} = constant$ . The coefficient of volume expansion of the gas is

A given quantity of an ideal gas is at pressure $p$ and absolute temperature $T$ . The isothermal bulk modulus of the gas is

An ideal gas is initially at temperature $T$ and volume $V$ . Its volume is increased by $Δ V$ due to an increase in temperature $Δ T$ , pressure remaining constant. The quantity $δ = Δ V / (V Δ T)$ varies with temperature as