Question

A  is placed in a cylinder and has a pressure of 33.3 kPa, a temperature of 25 oC, and a volume of 0.345 L. The pressure of the gas is increased to 40 kPa, and the volume is increased to 0.50 L. What is the temperature of the gas?

Answer 1

$P_1=33.3kPa$

$T_1=273+25=298k$

$V_1=0.345L$

$P_2=40kPa$

$T_2=?$

$V_2=0.50L$

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

Hence

$T_2=\frac{P_2V_2T_1}{P_1V_1}$

$T_2=\frac{40×0.50×298}{33.3×0.345}=518.78K$