Liquid helium at 4.2 K has a density of 0.147 g/mL. Suppose that a 1.00-L metal bottle that contains air at 118K and 1.0 atm pressure is sealed off. If we inject 120.0 mL of liquid helium and allow the entire system to warm to room temperature (25 °C), what is the pressure inside the bottle?
Add the partial pressures of the air and that of the vaporized helium.
120 mL of LHe has a mass of 0.147 g/ml x 120 mL = 17.64 g. That is = 4.406 moles. From that and the volume and temperature, you can calculate the partial pressure of He, using
P = nRT/V
R = 8.205*10^-2 atm*L/mole K
P(He)= 107.7 atm
The air partial pressure is much less. It starts out at 1.0 atm at 118K and rises to 1.0 x (298/118) = 2.5 atm at the higher temperature of 298 K.
Add the two partical pressures for the total pressure.
I don't guarantee my numbers, so check them yourself. The method should work.
thanks
You're welcome! Just be cautious when checking the numbers, as I tend to add a pinch of humor to my responses. Double-checking is always a good idea, especially when it comes to calculations. Have fun with your calculations and let me know if there's anything else I can help you with!
To calculate the pressure inside the bottle, we need to add the partial pressures of the air and the vaporized helium.
First, let's calculate the partial pressure of helium (P(He)):
Mass of liquid helium injected = density x volume
Mass of liquid helium injected = 0.147 g/mL x 120.0 mL = 17.64 g
Moles of liquid helium injected = mass / molar mass
Molar mass of helium (He) = 4.0026 g/mol
Moles of liquid helium injected = 17.64 g / 4.0026 g/mol = 4.406 mol
Next, we can use the ideal gas law to calculate the partial pressure of helium:
P(He) = (moles of helium x gas constant x temperature) / volume
Gas constant (R) = 8.205 * 10^-2 atm*L/mol*K
Temperature (T) = 25 °C = 298 K
Volume (V) = 1.00 L
P(He) = (4.406 mol x 8.205 * 10^-2 atm*L/mol*K x 298 K) / 1.00 L
P(He) ≈ 107.7 atm
Now, let's calculate the partial pressure of air (P(air)):
The air starts at 1.0 atm at 118 K. To find the pressure at the higher temperature of 298 K, we can use the equation:
P(air) = P(initial) x (T(final) / T(initial))
P(initial) = 1.0 atm
T(initial) = 118 K
T(final) = 298 K
P(air) = 1.0 atm x (298 K / 118 K)
P(air) ≈ 2.5 atm
Finally, we can find the total pressure inside the bottle by adding the partial pressures of helium and air:
Total pressure = P(He) + P(air)
Total pressure ≈ 107.7 atm + 2.5 atm
Total pressure ≈ 110.2 atm
Therefore, the pressure inside the bottle is approximately 110.2 atm.
To find the pressure inside the bottle after injecting liquid helium and allowing the system to warm to room temperature, we need to add the partial pressures of the air and the vaporized helium.
First, let's find the partial pressure of helium. The mass of 120 mL of liquid helium can be calculated by multiplying the density of helium (0.147 g/mL) by the volume injected (120 mL), which gives us 17.64 g. Converting this mass to moles, we obtain:
moles of helium = mass of helium / molar mass of helium = 17.64 g / 4.406 g/mol = 4.006 moles
Now, using the ideal gas equation:
P = nRT/V
where P is the pressure, n is the number of moles, R is the ideal gas constant (8.205x10^-2 atm*L/mol*K), T is the temperature in Kelvin, and V is the volume.
We know the number of moles of helium (4.006 moles), the temperature after warming the system to room temperature (25 °C or 298 K), and the volume of the bottle (1.00 L). Plugging in these values, we can calculate the partial pressure of helium:
P(He) = (4.006 moles) * (8.205x10^-2 atm*L/mol*K) * (298 K) / (1.00 L) = 107.7 atm
Next, let's find the partial pressure of the air. The air initially has a pressure of 1.0 atm at a temperature of 118 K. To find the pressure at the higher temperature of 298 K, we can use the ideal gas law:
P1/T1 = P2/T2
where P1 and P2 are the initial and final pressures, and T1 and T2 are the initial and final temperatures. Rearranging the equation to solve for the final pressure (P2), we have:
P2 = P1 * (T2 / T1) = 1.0 atm * (298 K / 118 K) = 2.5 atm
Finally, we can calculate the total pressure inside the bottle by adding the partial pressures of helium and air:
Total pressure = P(He) + P(air) = 107.7 atm + 2.5 atm = 110.2 atm
So, the pressure inside the bottle after injecting 120.0 mL of liquid helium and allowing the system to warm to room temperature is approximately 110.2 atm.