An empty LPG cylinder weight 14.8 kg. When full, it weight 29.0 kg and shows a pressure of 3.47 atm.

Q: Why do we subtract the empty cylinder's weight?

The pressure inside the cylinder is exerted *only* by the gas, not the metal container. The ideal gas law applies to the gas molecules, so we need the mass of the gas alone.

Q: Why is the volume considered constant?

An LPG cylinder is a rigid steel container. Even as gas is consumed, the remaining gas expands to fill the entire volume of the cylinder. Thus, the volume $V$ remains constant.

Q: Can we use the mass directly instead of moles in the ideal gas equation?

Yes! Since $n = \frac{m}{M}$, and the molar mass $M$ of LPG is constant, the number of moles is directly proportional to the mass. This allows us to use mass ratios directly.

Q: What if the temperature was not constant?

If the temperature changed from $T_1$ to $T_2$, you would use the combined relation $\frac{P_1}{m_1 T_1} = \frac{P_2}{m_2 T_2}$. But here, it says "ambient temperature", implying it remains constant.

Q: Why did the pressure decrease?

As gas is used, there are fewer gas molecules inside the rigid cylinder. Fewer molecules mean fewer collisions with the walls, which directly results in a lower pressure.

Question

An empty LPG cylinder weight $14.8	ext{ kg}$. When full, it weight $29.0	ext{ kg}$ and shows a pressure of $3.47	ext{ atm}$. In the course of use at ambient temperature, the mass of the cylinder is reduced to $23.0	ext{ kg}$. The final pressure inside of the cylinder is ......... $	ext{atm}$. (Nearest integer)
(Assume LPG of be an ideal gas)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a standard LPG cylinder in your kitchen. The problem gives us the weight of the cylinder in three different states: completely empty, completely full, and partially used.

The first trap to avoid is using the total weight of the cylinder in our gas equations. The pressure inside is exerted **only by the gas molecules**, not the heavy steel container!

Let's find the actual mass of the gas when the cylinder is full. We simply subtract the empty cylinder's weight from the full weight.

$$m_1 = 29.0 	ext{ kg} - 14.8 	ext{ kg} = 14.2 	ext{ kg}$$

At this initial state, the pressure $P_1$ is given as $3.47 	ext{ atm}$.

Now, after some gas is consumed, the total weight drops to $23.0 	ext{ kg}$. Let's find the new mass of the remaining gas.

$$m_2 = 23.0 	ext{ kg} - 14.8 	ext{ kg} = 8.2 	ext{ kg}$$

We need to find the new pressure $P_2$ corresponding to this remaining mass.

### The Master Equation

To connect the macroscopic properties of the gas, we bring in our trusty Ideal Gas Equation.

$$PV = nRT$$

We know that the number of moles $n$ is the given mass $m$ divided by the molar mass $M$. Let's substitute this into the equation.

$$PV = \frac{m}{M}RT$$

Now, let's look at our specific situation. The cylinder is a rigid steel container, so its volume $V$ is **strictly constant**. The problem states the process happens at "ambient temperature", meaning $T$ is constant. And of course, the molar mass $M$ of LPG doesn't change.

Since $V$, $T$, $R$, and $M$ are all constants, we can clearly see that the pressure is **directly proportional** to the mass of the gas.

$$P \propto m$$

### Final Calculation

Because of this direct proportionality, we can set up a simple ratio between the two states.

$$\frac{P_1}{P_2} = \frac{m_1}{m_2}$$

Let's carefully substitute the values we have found.

$$\frac{3.47}{P_2} = \frac{14.2}{8.2}$$

Now, it's just a matter of basic algebra to isolate $P_2$.

$$P_2 = \frac{3.47 	imes 8.2}{14.2}$$

Simplifying the fraction by dividing by $0.2$, we get:

$$P_2 = \frac{3.47 	imes 41}{71}$$

Calculating this value gives us:

$$P_2 \approx 2.003 	ext{ atm}$$

The question asks for the final pressure rounded to the nearest integer. Since $2.003$ is extremely close to $2$, our final answer is exactly $2 	ext{ atm}$. The pressure dropped proportionally as the gas was consumed, which perfectly aligns with our physical intuition!

Enter Numerical Value:

Visualized Solution

$m_{initial}$

$m_{final}$

$P V = n R T$

$\frac{P _{1}}{P _{2}} = \frac{m _{1}}{m _{2}}$

$P_{2}$

$Final Answer$

$What if T changes?$

The Sigma Insight: Gaseous State

Analyzing the Setup

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

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