Question

A 5.00 L sample of a gas at 25.0 °C is expanded under constant pressure to 12.5 L. What is the final temperature of the gas in °C?

Answer 1

Answer:
`472.22 \°C`

Explanation:

According to the Ideal Gas Law for an isobaric process (at a constant pressure):

`(V_(1))/(T_(1))=(V_(2))/(T_(2))` (1)

Where:

`V_(1)=5 L` is the initial volume of the sample

`T_(1)=25\°C + 273.15=298.15 K` is the initial temperature of the sample in Kelvin

`V_(2)=12.5 L` is the final volume of the sample

`T_(2)` is the final temperature of the sample

So, we have to find
`T_(2)` from (1):

`T_(2)=V_(2)(T_(1))/(V_(1))` (2)

`T_(2)=12.5 L(298.15 K)/(5 L)` (3)

`T_(2)=745.37 K` (4)

Transforming this result to Celsius:

`T_(2)=745.37 K-273.15=472.22 \°C` This is the final temperature in Celsius.