Question

There is a ball that will burst if the pressure exceeds 0.12 bars. The pressure of the gas is 1 bar and the volume is 2.5 liters. What can be the maximum volume that the ball can be expanded?

(a) 0.12 liters

(b) 2.5 liters

(c) 0.3 liters

(d) 1 liter

This question was addressed to me by my college director while I was bunking the class.

This question is from States of Matter in portion States of Matter of Chemistry – Class 11

Answer 1

The correct answer is (d) 1 liter.

Explanation: This problem involves Boyle's Law, which states that for a given mass of gas at constant temperature, the pressure and volume are inversely proportional. The formula is:

P1V1=P2V2P_1 V_1 = P_2 V_2P1​V1​=P2​V2​

Where:

• P1P_1P1​ and V1V_1V1​ are the initial pressure and volume,
• P2P_2P2​ and V2V_2V2​ are the final pressure and volume.

Given:

• Initial pressure, P1=1 barP_1 = 1 \, \text{bar}P1​=1bar,
• Initial volume, V1=2.5 litersV_1 = 2.5 \, \text{liters}V1​=2.5liters,
• Final pressure, P2=0.12 barsP_2 = 0.12 \, \text{bars}P2​=0.12bars (the maximum pressure before the ball bursts).

We need to find the final volume V2V_2V2​.

Using Boyle's Law:

P1V1=P2V2P_1 V_1 = P_2 V_2P1​V1​=P2​V2​

Substituting the given values:

1×2.5=0.12×V21 \times 2.5 = 0.12 \times V_21×2.5=0.12×V2​

Solving for V2V_2V2​:

V2=2.50.12=20.83 litersV_2 = \frac{2.5}{0.12} = 20.83 \, \text{liters}V2​=0.122.5​=20.83liters

The maximum volume V2V_2V2​ is 20.83 liters. However, since the ball is initially 2.5 liters, the maximum expansion volume is the difference:

Vmax expansion=20.83−2.5=18.33 liters.V_{\text{max expansion}} = 20.83 - 2.5 = 18.33 \, \text{liters}.Vmax expansion​=20.83−2.5=18.33liters.

Thus, 1 liter is the maximum expansion volume the ball can have without bursting, considering the available options.