If the pressure on an enclosed gas is doubled and its temperature rises from 27ºC to 127ºC, then its volume will

To solve this problem, we need to understand the relationship between pressure, volume, and temperature of a gas. The relevant law here is the **Combined Gas Law**, which is expressed as:

(P1 * V1) / T1 = (P2 * V2) / T2

Where:

• P1 is the initial pressure
• V1 is the initial volume
• T1 is the initial temperature in Kelvin
• P2 is the final pressure
• V2 is the final volume
• T2 is the final temperature in Kelvin

In the given problem:

• The initial temperature, T1, is 27ºC, which is 300K (since we convert to Kelvin by adding 273).
• The final temperature, T2, is 127ºC, which is 400K.
• The pressure is doubled, so P2 = 2 * P1.

Applying the Combined Gas Law:

(P1 * V1) / 300 = (2 * P1 * V2) / 400

Simplifying this equation:

V1/300 = 2V2/400

Multiply both sides by 400 to clear the fraction:

400 * V1 / 300 = 2 * V2

Which further simplifies to:

(4/3) * V1 = 2 * V2

Dividing both sides by 2:

(2/3) * V1 = V2

This shows that the final volume, V2, is **2/3 of the initial volume, V1**. Therefore, the volume of the gas will **decrease by 1/3**.