A sample of oxygen at 44◦C occupies 373 mL.
If this sample later occupies 1102 mL at 60◦C
and 1.8 atm, what was its original pressure?
Answer in units of atm
T1=44 °C=317 K
V1=373 ml=0.000373 m3
T2=60°C=333 K
V2=1102 ml=0.001102 m3
P2=1.8 atm=182385 Pa
R=8.31 J/(K mole)
P1 -?
"PV=nRT"
"n=\\frac{P_2V_2}{RT_2}=\\frac{182385x0.001102}{8.31x333}=0.0726 mole"
"P_1=\\frac{nRT_1}{V_1}=\\frac{0.0726x8.31x317}{0.000373}=512729 Pa"
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