# The volume of a gas is 10.2 L when the temperature is 9.5"^oC. If the volume is increased to 64.9 L without changing the pressure what is the new temperature?

Nov 20, 2016

The new temperature is $= 1524.5$ºC

#### Explanation:

we need to use Charles' law

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

The temperatures are in Kelvin (K)

K=ºC+273

${V}_{1} = 10.2 l$

T_1=9.5ºC=9.5+273=282.5K

${V}_{2} = 64.9 l$

so, ${T}_{2} = {V}_{2} \cdot {T}_{1} / {V}_{1} = 64.9 \cdot \frac{282.5}{10.2} = 1797.5 K$

T_2=1797.5-273=1524.5ºC