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

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Answer 1 · 2016-11-20T13:26:02

The new temperature is #=1524.5#ºC

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=10.2l#

#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#