# A gas is held at 3.8 atm and 500 K. If the pressure is then decreased to 1.2 atm, what will the new temperature be?

Mar 21, 2017

${T}_{2} = 157.89 K$

${T}_{2} = - {115.10}^{0} C$

#### Explanation:

By changing the physical conditions like pressure, temperature and volume of the gas, the moles of the gas remain constant.
${n}_{1} = {n}_{2}$

According to the ideal gas equation,

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

The volume of the process is also constant . As, according to the Avogadro's law,
$V \alpha n$

So, ${V}_{1} = {V}_{2}$

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

$\frac{3.8}{500} = \frac{1.2}{T} _ 2$

${T}_{2} = \frac{1.2}{3.8} \times 500$

${T}_{2} = 157.89 K$

${T}_{2} = - {115.10}^{0} C$

So, the new temperature will be $2280 K$.