# 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?

Nov 29, 2015

The new temperature will be $\text{270. K}$, 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: ${P}_{1} = \text{4.882 atm}$
temperature 1: ${T}_{1} = {8}^{\text{o""C"+273.15="281 K}}$
pressure 2: ${P}_{2} = \text{4.690 atm}$

Unknown
temperature 2: ${T}_{2}$

Solution
Rearrange the equation to isolate ${T}_{2}$ and solve.

${T}_{2} = \frac{{T}_{1} {P}_{2}}{P} _ 1$

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