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):  

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$ (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):

$T_{2}=V_{2}\frac{T_{1}}{V_{1}}$ (2)

$T_{2}=12.5 L\frac{298.15 K}{5 L}$ (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.