Question
A sample of oxygen gas occupies 500.0 mL at -185°C and 75.0 cmHg. Calculate the temperature in °C if the gas has a volume of 221.0 mL at a 55.0 cmHg.
Answers
PV/T remains constant, so
(75.0)(500)/(-185+271) = (55)(221.0)/T
T = 28°K
Note that we don't really have to convert to L and Pa, because the same conversion factors would be applied to both sides of the equation. The only wrinkle is the °C-°K conversion, because that is additive, not multiplicative.
(75.0)(500)/(-185+271) = (55)(221.0)/T
T = 28°K
Note that we don't really have to convert to L and Pa, because the same conversion factors would be applied to both sides of the equation. The only wrinkle is the °C-°K conversion, because that is additive, not multiplicative.
This answer is not correct
Steve gave you the combined gas law, but it seems that you were confused. The gas law is as followed:
P1V1/T1=P2V2/T2
Where
P1=75.0 cmHg
V1=500.0 mL
T1=273K +(-185ºC)=88K
P2=55.0 cmHg
V2=221.0mL
and
T2=?
Solve for T2:
T2=[T1*(P2V2)]/P1V1
T2=[88K*(55.0 cmHg*221.0mL)/(75.0 cmHg*500.0 mL)
T2=28.6 K
Convert Kelvin to Celsius:
K=273K +C
K-273=C
28.6-273=-244 º C <== Three significant figures
P1V1/T1=P2V2/T2
Where
P1=75.0 cmHg
V1=500.0 mL
T1=273K +(-185ºC)=88K
P2=55.0 cmHg
V2=221.0mL
and
T2=?
Solve for T2:
T2=[T1*(P2V2)]/P1V1
T2=[88K*(55.0 cmHg*221.0mL)/(75.0 cmHg*500.0 mL)
T2=28.6 K
Convert Kelvin to Celsius:
K=273K +C
K-273=C
28.6-273=-244 º C <== Three significant figures