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

1 Answer
Nov 20, 2016

The new temperature is =1524.5=1524.5ºC

Explanation:

we need to use Charles' law

V_1/T_1=V_2/T_2V1T1=V2T2

The temperatures are in Kelvin (K)

K=ºC+273

V_1=10.2l

T_1=9.5ºC=9.5+273=282.5K

V_2=64.9l

so, T_2=V_2*T_1/V_1=64.9*282.5/10.2=1797.5K

T_2=1797.5-273=1524.5ºC