The volume of a sample of oxygen is 300.0mL when the pressure is 1.00 atm and the temperature is 27.0 degrees C. At what temperature is the volume 1.00 L and the pressure 0.500 atm?
To solve this problem, we can use the combined gas law, which relates the initial and final volumes, pressures, and temperatures of a gas sample.
The combined gas law equation is given by:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
Let's plug in the given values into the equation and solve for T2.
Given:
P1 = 1.00 atm
V1 = 300.0 mL = 0.300 L (since 1 L = 1000 mL)
T1 = 27.0 degrees C = 300.0 K (since temperature must be in Kelvin)
P2 = 0.500 atm
V2 = 1.00 L
Now, substitute the given values into the equation:
(1.00 atm * 0.300 L) / (300.0 K) = (0.500 atm * 1.00 L) / (T2)
Simplifying the equation:
0.00333 L/atm/K = 0.5 L/atm / T2
To solve for T2, rearrange the equation:
T2 = 0.5 L/atm / (0.00333 L/atm/K)
T2 = 150.15 K
Therefore, at a volume of 1.00 L and a pressure of 0.500 atm, the temperature is approximately 150.15 K.
To solve this problem, we can use the combined gas law, which relates the volume, pressure, and temperature of a gas.
The formula for the combined gas law is:
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
Where:
P₁ and P₂ are the initial and final pressures,
V₁ and V₂ are the initial and final volumes,
T₁ and T₂ are the initial and final temperatures.
Let's plug in the given values into the equation:
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
We have:
P₁ = 1.00 atm
V₁ = 300.0 mL = 0.3000 L
T₁ = 27.0 °C = 27.0 + 273.15 K (converting to Kelvin)
P₂ = 0.500 atm
V₂ = 1.00 L
T₂ = unknown (we need to solve for this)
Substituting these values into the equation:
(1.00 atm * 0.3000 L) / (27.0 + 273.15 K) = (0.500 atm * 1.00 L) / T₂
Now we can solve for T₂. Let's rearrange the equation to isolate T₂:
T₂ = ((0.500 atm * 1.00 L) / (1.00 atm * 0.3000 L)) * (27.0 + 273.15 K)
Calculating this:
T₂ = (0.500 / 0.3000) * (27.0 + 273.15) K
T₂ = 0.500 * 300.15
T₂ ≈ 150.07 K
Therefore, at a volume of 1.00 L and a pressure of 0.500 atm, the temperature would be approximately 150.07 K.