# The pressure of a sample of argon at 20°C is decreased from 720 mmHg to 360 mmHg. What was the final temperature of the argon gas?

Jun 25, 2016

${T}_{2} = 146.5 K$

or

${T}_{2} = - {126.5}^{o} C$

#### Explanation:

For this solution we will use the Gay-Lussac Law relating pressure and temperature in gases. ${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

For this case the values are

${P}_{1} = 720 m m H g$
${T}_{1} = {20}^{o} C \mathmr{and} 293 K$
${P}_{2} = 360 m m H g$
T_1 = ?K

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

$\frac{720 m m H g}{293 K} = \frac{360 m m H g}{T} _ 2$

We can invert both sides to put the missing variable in the numerator.

$\frac{293 K}{720 m m H g} = {T}_{2} / \left(360 m m H g\right)$

$\frac{293 K}{720 \cancel{m m H g}} \left(360 \cancel{m m H g}\right) = {T}_{2} / \left(\cancel{360 m m H g}\right) \left(\cancel{360 m m H g}\right)$

${T}_{2} = 146.5 K$

or

${T}_{2} = - {126.5}^{o} C$