# If a sample of N gas was put into a flexible container at an pressure of 25.5kPa, temperature of 35.9 C and a volume of 500. mL, what would the final temperature of the gas be if the pressure was changed to .625 atm and the volume was change to 625mL?

Mar 5, 2017

${705}^{\circ} \text{C}$

#### Explanation:

First look at the things that need to be converted

${35.9}^{\circ} \text{C" = "(35.9 + 273.15) K" = "309.0 K}$

$\text{25.5 kPa = 0.2467 atm}$

$\text{500 mL = 0.500 L}$

$\text{625 mL = 0.625 L}$

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

Plug in the variables

$\frac{\text{0.2467 atm" xx "0.500 L")/("309.0 K") = ("0.625 atm" xx "0.625 L}}{T} _ 2$

${T}_{2} = 978.5 K$

or ${T}_{2} = {\left(978.5 - 273.15\right)}^{\circ} \text{C" = 705 ^@"C}$