A 6.35 L sample of carbon dioxide is collected at 550 ° C and 90.312 kPa. What volume will the gas occupy at 106.391 kPa and 20.0 ° C?
P1×V1T1=P2×V2T2=nRV2=P1×V1×T2T1×P2=n×R×T2P2=90.312×6.35×293773×106.391=2.043L\frac{P_1×V_1}{T_1}=\frac{P_2×V_2}{T_2}=nR\\V_2=\frac{P_1×V_1×T_2}{T_1×P_2}=\frac{n×R×T_2}{P_2}\\=\frac{90.312×6.35×293}{773×106.391}=2.043L \\ \\T1P1×V1=T2P2×V2=nRV2=T1×P2P1×V1×T2=P2n×R×T2=773×106.39190.312×6.35×293=2.043L
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