A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of neg

Q: Why do we multiply pressure by area to get force?

Pressure is defined as force per unit area ($p = F/A$). Therefore, the force exerted by a gas on a surface is the pressure multiplied by the area of that surface ($F = pA$).

Q: Why is the temperature $T$ the same for both parts?

The problem explicitly states that 'Each part of the cylinder contains $n$ moles of an ideal gas at equal temperature $T$'. We just follow the given conditions.

Q: What happens if the cylinder is horizontal instead of vertical?

If the cylinder were horizontal, gravity would act perpendicular to the motion of the piston. In that case, the equilibrium condition would simply be $p_1 A = p_2 A$, meaning the pressures on both sides would be equal.

Q: Why did the area $A$ cancel out?

The area $A$ cancels out because the volume is proportional to the area ($V = Al$), and the force is also proportional to the area ($F = pA$). When you substitute $p = nRT / (Al)$ into $F = pA$, the $A$ in the numerator and denominator cancel each other.

Q: Can $l_2$ be greater than $l_1$?

The problem states $l_1 > l_2$. If $l_2$ were greater than $l_1$, the term $(l_1 - l_2)$ would be negative, implying a negative mass, which is physically impossible in this setup. The higher pressure must be at the bottom to support the weight of the piston.

Question

A vertical closed cylinder is separated into two parts by a frictionless piston of mass $m$ and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the piston is $l_1$ and that below the piston is $l_2$, such that $l_1 > l_2$. Each part of the cylinder contains $n$ moles of an ideal gas at equal temperature $T$. If the piston is stationary, its mass $m$, will be given by (where, $R$ is universal gas constant and $g$ is the acceleration due to gravity)

Options:

(A) $\frac{nRT}{g} \left[ \frac{l_1 - l_2}{l_1 l_2} \right]$

(B) $\frac{nRT}{g} \left[ \frac{1}{l_2} + \frac{1}{l_1} \right]$

(C) $\frac{RT}{g} \left[ \frac{2l_1 + l_2}{l_1 l_2} \right]$

(D) $\frac{RT}{ng} \left[ \frac{l_1 - 3l_2}{l_1 l_2} \right]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a vertical cylinder divided into two compartments by a movable piston of mass $m$. The gas in the upper compartment has a volume $V_1 = A l_1$, and the gas in the lower compartment has a volume $V_2 = A l_2$, where $A$ is the cross-sectional area of the cylinder. Both compartments contain $n$ moles of an ideal gas at the same temperature $T$.

### The Master Equation: Equilibrium
The key to solving this problem lies in the word **stationary**. Since the piston is not moving, it must be in mechanical equilibrium. This means the net force acting on it is zero. Let's break down the forces:
1. The gas above pushes down with a force $p_1 A$.
2. Gravity pulls the piston down with a force $mg$.
3. The gas below pushes up with a force $p_2 A$.

Balancing these forces, we get:
$$p_1 A + mg = p_2 A$$

Rearranging to solve for the weight of the piston:
$$mg = (p_2 - p_1) A$$

### Applying the Ideal Gas Law
We don't know the pressures $p_1$ and $p_2$ directly, but we do know the number of moles, temperature, and volume. This is where the Ideal Gas Law, $pV = nRT$, comes to the rescue!

We can express the pressures as:
$$p_1 = \frac{nRT}{V_1} = \frac{nRT}{A l_1}$$
$$p_2 = \frac{nRT}{V_2} = \frac{nRT}{A l_2}$$

### Final Calculation
Now, let's substitute these pressure expressions back into our equilibrium equation:
$$mg = \left( \frac{nRT}{A l_2} - \frac{nRT}{A l_1} \right) A$$

Notice how the area $A$ is common in the denominator and cancels out perfectly with the $A$ outside the bracket. This is a beautiful moment in physics where an unknown geometric parameter simply vanishes!
$$mg = nRT \left( \frac{1}{l_2} - \frac{1}{l_1} \right)$$

Taking the common denominator for the terms inside the bracket:
$$mg = nRT \left( \frac{l_1 - l_2}{l_1 l_2} \right)$$

Finally, isolating the mass $m$:
$$m = \frac{nRT}{g} \left[ \frac{l_1 - l_2}{l_1 l_2} \right]$$

And there we have it! The mass of the piston is elegantly expressed in terms of the given parameters.

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Visualized Solution

$F r ee B o d y D ia g r am$

$E q u i l ib r i u m C o n d i t i o n$

$R e a r r an g in g f or m g$

$I d e a l G a s E q u a t i o n$

$S u b s t i t u t in g P r ess u r es$

$S im pl i f y in g t h e E x p r ess i o n$

$F ina l M a ss E x p r ess i o n$

The Sigma Insight: Kinetic Theory of Gases

Analyzing the Setup Imagine a vertical cylinder divided into two compartments by a movable piston of mass $m$

The Master Equation

Applying the Ideal Gas Law We don't know the pressures $p_{1}$ and $p_{2}$ directly, but we do know the number of moles, temperature, and volume

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Free Body Diagram

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equilibrium Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rearranging for mg

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ideal Gas Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Pressures

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Mass Expression