A 71.0 L cylinder containing CH2O at a pressure of 0.550 atm is connected by a valve to 1.50 L cylinder containing N2 at 266.0 torr pressure. Calculate the partial pressure (atm) of N2 when the valve is opened.
What's the big hang up on this problem? Surely you know how to do parts of it.
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To calculate the partial pressure of N2 when the valve is opened, we need to use the ideal gas law equation: PV = nRT.
First, let's convert the pressure of the N2 in the second cylinder to atm:
266.0 torr = 266.0/760 atm ≈ 0.350 atm
Next, let's calculate the number of moles of CH2O in the first cylinder:
Since the volume is given as 71.0 L and the pressure is given as 0.550 atm, we can rewrite the ideal gas law equation as n = PV/RT.
nCH2O = (0.550 atm)(71.0 L) / (0.0821 L·atm/(mol·K))(298 K)
≈ 16.50 mol
Now, let's determine the number of moles of N2 initially present in the second cylinder using the ideal gas law:
Since the volume is given as 1.50 L and the pressure is given as 0.350 atm, we can rewrite the ideal gas law equation as n = PV/RT.
nN2 = (0.350 atm)(1.50 L) / (0.0821 L·atm/(mol·K))(298 K)
≈ 0.0254 mol
When the valve is opened, the N2 from the second cylinder will mix with the CH2O in the first cylinder, resulting in a total volume of 72.50 L (71.0 L + 1.50 L).
To calculate the partial pressure of N2 after the valve is opened, we use the equation:
P₂ = (n₂ × R × T) / V₂,
where P₂ is the partial pressure of N2, n₂ is the number of moles of N2, R is the ideal gas constant, T is the temperature, and V₂ is the total volume.
P₂ = (0.0254 mol × 0.0821 L·atm/(mol·K) × 298 K) / 72.50 L
≈ 0.0111 atm
Therefore, the partial pressure of N2 when the valve is opened is approximately 0.0111 atm.