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.
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.
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
To calculate the temperature in °C, we can use the combined gas law equation:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
where:
P₁ and P₂ are the initial and final pressure,
V₁ and V₂ are the initial and final volume, and
T₁ and T₂ are the initial and final temperature.
Let's first convert the temperatures to Kelvin (K). The Kelvin scale is obtained by adding 273.15 to the Celsius temperature.
Given:
V₁ = 500.0 mL
T₁ = -185°C + 273.15 = 88.15 K
P₁ = 75.0 cmHg
V₂ = 221.0 mL
P₂ = 55.0 cmHg
Now, we can rearrange the equation and solve for T₂:
T₂ = (P₂ * V₂ * T₁) / (P₁ * V₁)
Substituting the given values:
T₂ = (55.0 cmHg * 221.0 mL * 88.15 K) / (75.0 cmHg * 500.0 mL)
Now, let's calculate the temperature:
T₂ = (55.0 * 221.0 * 88.15) / (75.0 * 500.0)
T₂ ≈ 60.79 K
Finally, let's convert the temperature back to °C:
T₂ ≈ 60.79 K - 273.15 = -212.36°C
Therefore, the temperature in °C when the gas has a volume of 221.0 mL at a pressure of 55.0 cmHg is approximately -212.36°C.