Posted by **Harry** on Friday, February 14, 2014 at 6:10pm.

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.

- Chemistry -
**Steve**, Friday, February 14, 2014 at 6:31pm
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.

- Chemistry -
**Harry**, Friday, February 14, 2014 at 6:40pm
This answer is not correct

- Chemistry -
**Devron**, Friday, February 14, 2014 at 7:14pm
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

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