A 1.25-g sample of CO2 is contained in a 750.-mL flask at 22.5 °C. What is the pressure
of the gas?
m=1.25 gV=750 mL=0.75 LT=22.5+273.15=295.65 KM(CO2)=44.01 g/moln(CO2)=1.2544.01=0.0284 molm=1.25 \;g \\ V=750 \;mL = 0.75 \;L \\ T= 22.5+273.15 = 295.65 \;K \\ M(CO_2) = 44.01 \;g/mol \\ n(CO_2) = \frac{1.25}{44.01} = 0.0284 \;molm=1.25gV=750mL=0.75LT=22.5+273.15=295.65KM(CO2)=44.01g/moln(CO2)=44.011.25=0.0284mol
Ideal gas law
pV = nRT
R = 0.08206 L×atm/mol×K
p=nRTVp=0.0284×0.08206×295.650.75=0.918 atmp=\frac{nRT}{V} \\ p= \frac{0.0284 \times 0.08206 \times 295.65}{0.75} = 0.918 \;atmp=VnRTp=0.750.0284×0.08206×295.65=0.918atm
Answer: 0.918 atm
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