A sample of nitrogen, N2, occupies 45.0 mL at 27 °C and 600 torr. What pressure will it have if cooled to –73 °C while the volume remains constant?
V=constant=45.0 mLT1=27+273=300 Kp1=600 torrT2=−73+273=200 Kp1T1=p2T2p2=p1T2T1=600×200300=400 torrV = constant = 45.0 \;mL \\ T_1 = 27+273 = 300 \;K \\ p_1 = 600 \; torr \\ T_2 = -73 + 273 = 200 \;K \\ \frac{p_1}{T_1} = \frac{p_2}{T_2} \\ p_2 = \frac{p_1 T_2}{T_1} \\ = \frac{600 \times 200}{300} = 400 \; torrV=constant=45.0mLT1=27+273=300Kp1=600torrT2=−73+273=200KT1p1=T2p2p2=T1p1T2=300600×200=400torr
Answer: 400 torr
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