What is the volume of a sample of ethane at 467k and 1.1atm ,if it occupies 405ml at 298k and 1.1atm
According to the combined gas law,
P1V1T1=P2V2T2\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}T1P1V1=T2P2V2
Therefore,
V2=P1V1T2T1P2=1.1 atm×405 mL×467 K298 K×1.1 atm=635 mLV_2=\frac{P_1V_1T_2}{T_1P_2}=\frac{1.1\ atm\times405\ mL\times467\ K}{298\ K\times1.1\ atm}=635\ mLV2=T1P2P1V1T2=298 K×1.1 atm1.1 atm×405 mL×467 K=635 mL
Answer: 635 mL
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