An LPG cylinder contains gas at a pressure of 300 kPa at 27°C. The cylinder can withstand the pressu

Q: Why can't we use the temperature in Celsius directly in the formula?

Gas laws are derived based on the absolute zero of temperature, where molecular motion theoretically stops. The Kelvin scale starts at this absolute zero, making pressure directly proportional to temperature. Celsius is a relative scale, so using it would break the direct proportionality.

Q: How do we know the volume is constant?

The problem states the gas is in an 'LPG cylinder.' Cylinders are made of rigid steel and do not expand significantly under normal or even fire conditions until they structurally fail (burst). Therefore, we assume $V$ is constant.

Q: What if the question didn't have a typo and we used $12 \times 10^6\text{ Pa}$?

If you used $12 \times 10^6\text{ Pa}$, your final temperature would be $12000\text{ K}$, which is $11727^\circ\text{C}$. While mathematically correct for those numbers, it doesn't match the official answer of $927^\circ\text{C}$, confirming the typo in the original paper.

Q: Why did we convert $300\text{ kPa}$ to $300 \times 10^3\text{ Pa}$?

To use a ratio like $p_1/p_2$, both pressures must be in the exact same units. Since the bursting pressure was given in Pascals ($\text{Pa}$), we converted kilopascals ($\text{kPa}$) to Pascals by multiplying by $10^3$.

Q: Does the type of gas (LPG) affect the calculation?

No! At high temperatures and relatively low pressures, most real gases behave like ideal gases. The ideal gas law ($pV = nRT$) and its derivatives apply regardless of the gas's chemical identity.

Question

An LPG cylinder contains gas at a pressure of $300	ext{ kPa}$ at $27^\circ	ext{C}$. The cylinder can withstand the pressure of $12 	imes 10^6	ext{ Pa}$. The room in which the cylinder is kept catches fire. The minimum temperature at which the bursting of cylinder will take place is ……… $^\circ	ext{C}$. (Nearest integer)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine an LPG cylinder sitting quietly in a room. It is a rigid, steel container.

This rigidity is the most crucial physical constraint of our problem. It means that no matter how much the temperature or pressure changes, the **volume of the gas remains absolutely constant**.

Initially, the gas inside is at a pressure of $300	ext{ kPa}$ and a normal room temperature of $27^\circ	ext{C}$.

Suddenly, a fire breaks out! As the temperature of the room skyrockets, the kinetic energy of the gas molecules inside the cylinder increases drastically.

They start colliding with the walls of the cylinder with much greater force and frequency. Because the walls cannot expand to relieve this stress, the pressure builds up rapidly.

The cylinder has a structural limit; it can only withstand a maximum pressure before it catastrophically bursts.

### The Typo and The Setup

Before we dive into the math, we need to address a critical detail in the problem statement.

The question states the bursting pressure is $12 	imes 10^6	ext{ Pa}$. However, based on the official JEE answer key and standard physical parameters, this is a known typographical error in the exam paper.

The intended bursting pressure is actually **$1.2 	imes 10^6	ext{ Pa}$**. We will proceed with this corrected value to arrive at the intended solution.

Next, we must prepare our data. In thermodynamics, we can never use Celsius directly because it is a relative scale.

We must convert our initial temperature to the absolute Kelvin scale.

$$T_1 = 27 + 273 = 300	ext{ K}$$

We also need to ensure our pressure units match. Let's convert the initial pressure from kilopascals to Pascals.

$$p_1 = 300 	imes 10^3	ext{ Pa}$$

### The Master Equation

Since the volume and the amount of gas are constant, we rely on **Gay-Lussac's Law**.

This law states that the pressure of a fixed mass of gas is directly proportional to its absolute temperature.

Mathematically, this gives us our master equation:

$$\frac{p_1}{T_1} = \frac{p_2}{T_2}$$

Now, we substitute our known values into this elegant relationship.

$$\frac{300 	imes 10^3}{300} = \frac{1.2 	imes 10^6}{T_2}$$

### Final Calculation

Take a close look at the left side of our equation. It simplifies beautifully!

The $300$ in the numerator and the denominator cancel out perfectly.

$$10^3 = \frac{1.2 	imes 10^6}{T_2}$$

This makes our algebraic manipulation incredibly straightforward. We rearrange the equation to solve for our unknown bursting temperature, $T_2$.

$$T_2 = \frac{1.2 	imes 10^6}{10^3}$$

By subtracting the exponents, we find the temperature in Kelvin.

$$T_2 = 1.2 	imes 10^3	ext{ K} = 1200	ext{ K}$$

We have the bursting temperature, but the question specifically asks for the answer in degrees Celsius.

We must convert it back by subtracting $273$.

$$T_2 = 1200 - 273 = 927^\circ	ext{C}$$

The minimum temperature at which the cylinder will burst is **$927^\circ	ext{C}$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Gay-Lussac's Law

Unit Conversions

Substituting Values

Simplifying the Equation

Solving for $T_{2}$

Converting to Celsius

Alternative Scenarios

The Sigma Insight: Gaseous State

Analyzing the Setup

The Typo and The Setup

The Master Equation

Final Calculation

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