Question

There is a balloon filled with a gas at 26-degree centigrade and has a volume of about 2 liters when the balloon is taken to a place which is at 39-degree centigrade, what would be the volume of the gas that is inside the balloon?

(a) 2 liters

(b) 3 liters

(c) 1.5 liters

(d) 0.67 liters

The question was asked in a job interview.

Asked question is from States of Matter in chapter States of Matter of Chemistry – Class 11

Answer 1

To solve this, we can use Charles's Law, which states that for a given mass of gas at constant pressure, the volume of the gas is directly proportional to its temperature (in Kelvin). The formula is:

V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}T1​V1​​=T2​V2​​

Where:

• V1V_1V1​ is the initial volume,
• T1T_1T1​ is the initial temperature (in Kelvin),
• V2V_2V2​ is the final volume,
• T2T_2T2​ is the final temperature (in Kelvin).

Given:

• V1=2 litersV_1 = 2 \, \text{liters}V1​=2liters,
• Initial temperature T1=26∘C=26+273=299 KT_1 = 26^\circ \text{C} = 26 + 273 = 299 \, \text{K}T1​=26∘C=26+273=299K,
• Final temperature T2=39∘C=39+273=312 KT_2 = 39^\circ \text{C} = 39 + 273 = 312 \, \text{K}T2​=39∘C=39+273=312K.

We need to find V2V_2V2​, the final volume.

Applying Charles's Law:

V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}T1​V1​​=T2​V2​​

Substitute the known values:

2299=V2312\frac{2}{299} = \frac{V_2}{312}2992​=312V2​​

Solving for V2V_2V2​:

V2=2×312299=2.09 litersV_2 = \frac{2 \times 312}{299} = 2.09 \, \text{liters}V2​=2992×312​=2.09liters

Thus, the volume of the gas at 39°C is approximately 2.1 liters. Since this value is not exactly one of the choices, we can round it to 2 liters for practical purposes.

Answer: (a) 2 liters.