Question

When the pressure that a gas exerts on a sealed container changes from 53.0 Bar to 35.2 Bar, the temperature changes from _____ C to 12.0 C. Use P1/T1=P2/T2

Answer 1

Answer:

$156.1^{\circ}C$

Explanation:

We can solve the problem by using Gay-Lussac's law, which states that:

"For a fixed mass of ideal gas kept at constant volume, the pressure of the gas is directly proportional to its absolute temperature"

Mathematically:

$\frac{p}{T}=const.$

where

p is the pressure

T is the absolute temperature

The equation can be also rewritten as

$\frac{p_1}{T_1}=\frac{p_2}{T_2}$

where, in this problem:

$p_1 = 53.0 Bar$ is the initial pressure

$p_2=35.2 Bar$ is the final pressure

$T_2=12.0^{\circ}+273=285 K$ is the final temperature

Solving for T1, we find the initial temperature:

$T_1=\frac{p_1 T_2}{p_2}=\frac{(53.0)(285)}{35.2}=429.1 K$

Converted into Celsius degrees,

$T_2=429.1-273=156.1^{\circ}C$