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?
P1=33.3kPaP_1=33.3kPaP1=33.3kPa
T1=273+25=298kT_1=273+25=298kT1=273+25=298k
V1=0.345LV_1=0.345LV1=0.345L
P2=40kPaP_2=40kPaP2=40kPa
T2=?T_2=?T2=?
V2=0.50LV_2=0.50LV2=0.50L
P1V1T1=P2V2T2\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}T1P1V1=T2P2V2
Hence
T2=P2V2T1P1V1T_2=\frac{P_2V_2T_1}{P_1V_1}T2=P1V1P2V2T1
T2=40×0.50×29833.3×0.345=518.78KT_2=\frac{40×0.50×298}{33.3×0.345}=518.78KT2=33.3×0.34540×0.50×298=518.78K
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