A sample of oxygen occupies 15.0 mL under a pressure of 0.928 atm at 25.0 C. What would be its volume at another time when the pressure was 0.948 atm and the temperature 19.0 C?
Ah, a chemistry question! Let me put on my lab coat and goggles. Now, to calculate the new volume of oxygen, we can use the combined gas law equation:
(P1 x V1) / (T1) = (P2 x V2) / (T2)
So, let's plug in the values we have:
(0.928 atm x 15.0 mL) / (25.0 C) = (0.948 atm x V2) / (19.0 C)
Now, let me do some quick calculations in my funny-looking clown calculator:
(0.928 atm x 15.0 mL x 19.0 C) / (0.948 atm) = V2
And the result is... ta-da!
V2 = 16.10 mL
So, the volume of oxygen at the new time would be approximately 16.10 mL. Just don't let the oxygen get too inflated with pride!
To solve this problem, we can use the ideal gas law, which states:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal gas constant
T = Temperature
To find the volume of the oxygen at another time, we can use the ratio between the initial and final conditions.
Step 1: Convert the initial conditions to Kelvin
To convert from Celsius to Kelvin, add 273.15:
Initial temperature (T1) = 25.0 + 273.15 = 298.15 K
Step 2: Convert the final conditions to Kelvin
Final temperature (T2) = 19.0 + 273.15 = 292.15 K
Step 3: Determine the ratio of pressures (P1/P2)
Ratio of pressures (P1/P2) = 0.928 atm / 0.948 atm ≈ 0.9772
Step 4: Determine the ratio of temperatures (T1/T2)
Ratio of temperatures (T1/T2) = 298.15 K / 292.15 K ≈ 1.0205
Step 5: Calculate the ratio of volumes (V1/V2)
Using the combined gas law, we can rearrange the formula for volume:
V1/V2 = (P1/P2) x (T2/T1)
V1/V2 = 0.9772 x 1.0205 ≈ 0.9977
Step 6: Solve for V2 (final volume)
V2 = (V1 / (V1/V2)) = 15.0 mL / 0.9977 ≈ 15.04 mL
Therefore, the volume of oxygen at another time when the pressure was 0.948 atm and the temperature was 19.0 °C would be approximately 15.04 mL.
To solve this problem, we can use the combined gas law equation, which combines Boyle's Law, Charles's Law, and Gay-Lussac's Law. The equation is as follows:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
Where:
P₁ and P₂ are the initial and final pressures, respectively.
V₁ and V₂ are the initial and final volumes, respectively.
T₁ and T₂ are the initial and final temperatures, measured in Kelvin.
Let's start by converting the given temperatures from Celsius to Kelvin using the formula:
T(K) = T(°C) + 273.15
The initial temperature, T₁ = 25.0°C + 273.15 = 298.15 K
The final temperature, T₂ = 19.0°C + 273.15 = 292.15 K
Now we can substitute the given values into the combined gas law equation and solve for V₂:
(0.928 atm * 15.0 mL) / (298.15 K) = (0.948 atm * V₂) / (292.15 K)
To find V₂, we can rearrange the equation:
V₂ = (0.928 atm * 15.0 mL * 292.15 K) / (0.948 atm * 298.15 K)
By plugging in the values and performing the calculation:
V₂ ≈ 14.390 mL
Therefore, the volume of oxygen at another time, when the pressure is 0.948 atm and the temperature is 19.0°C, would be approximately 14.390 mL.