Answer to Question #218914 in General Chemistry for Isabella

Question #218914

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?


1
Expert's answer
2021-07-20T11:29:01-0400

P1=33.3kPaP_1=33.3kPa

T1=273+25=298kT_1=273+25=298k

V1=0.345LV_1=0.345L



P2=40kPaP_2=40kPa

T2=?T_2=?

V2=0.50LV_2=0.50L


P1V1T1=P2V2T2\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}


Hence


T2=P2V2T1P1V1T_2=\frac{P_2V_2T_1}{P_1V_1}


T2=40×0.50×29833.3×0.345=518.78KT_2=\frac{40×0.50×298}{33.3×0.345}=518.78K


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