Question

A compressed gas has a pressure of 4.882 atm at a temperature of 8° C. The next day, the same container of gas has a pressure of 4.690 atm. What is the new temperature?

Answer 1

The new temperature will be `"270. K"`, with three significant figures.

Explanation:

This is an example of Gay-Lussac's law, which states that when volume is kept constant, the pressure is directly proportional to the temperature in Kelvins.

The equation is `P_1/T_1=P_2/T_2`

Given/Known
pressure 1: `P_1="4.882 atm"`
temperature 1: `T_1=8^"o""C"+273.15="281 K"`
pressure 2: `P_2="4.690 atm"`

Unknown
temperature 2: `T_2`

Solution
Rearrange the equation to isolate `T_2` and solve.

`T_2=(T_1P_2)/P_1`

`T_2=((281"K"xx4.690"atm"))/(4.882"atm")="270. K"` (rounded to three significant figures)