I need help with these two problems:
At a given pressure and temperature, it takes 4.55 min for a 1.5 L sample of He to effuse through a membrane. How long does it take for 1.5 L of H2 to effuse under the same conditions?
A gas-filled weather balloon with a volume of 58.7 L is released at sea-level conditions of 755 torr and 23°C. The balloon can expand to a maximum volume of 788 L. When the balloon rises to an altitude at which the temperature is -5°C and the pressure is 0.051 atm, will it reach its maximum volume?
I tried working this one out by following this equation T2*P1*V1/(T1*P2)*(1/760)= (268K*755 torr*58.7L)/(296K*0.051atm)*(1/760)=1035.25 L, but the answer is somehow wrong when I put it into the computer.
On the first, the law of diffusion is used: Effusion is inversely related to mole mass..
rateHe/rateH2= sqrt ( molmassH2)/sqrt(molemassH2)
in your case, rates is volume/time
The problem with the second problem may be that what you are to enter is a yes or no answer and not the volume you have calculated.
To solve the first problem, you can use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
1. First, we need to calculate the molar mass of each gas. The molar mass of Helium (He) is approximately 4 g/mol, and the molar mass of Hydrogen (H2) is approximately 2 g/mol.
2. Now, let's set up the equation using Graham's law:
Rate of effusion (He) / Rate of effusion (H2) = √(molar mass (H2) / molar mass (He))
Let's assign the given values:
Rate of effusion (He) = 1.5 L / 4.55 min (given)
Molar mass (He) = 4 g/mol
Molar mass (H2) = 2 g/mol
Plugging in the values:
1.5 L / 4.55 min / Rate of effusion (H2) = √(2 g/mol / 4 g/mol)
Rearranging the equation to solve for Rate of effusion (H2):
Rate of effusion (H2) = (1.5 L / 4.55 min) * √(4 g/mol / 2 g/mol)
Simplifying the expression:
Rate of effusion (H2) = (1.5 / 4.55) * √2 ≈ 0.515 L/min
Therefore, it will take approximately 0.515 minutes for 1.5 L of H2 to effuse under the same conditions.
Now, let's move on to the second problem.
To solve the second problem, you can use the combined gas law, which relates the pressure, volume, and temperature of an ideal gas.
1. Assign the given values:
Initial conditions:
P1 (initial pressure) = 755 torr
V1 (initial volume) = 58.7 L
T1 (initial temperature) = 23°C = 23 + 273 = 296 K
Final conditions:
P2 (final pressure) = 0.051 atm
T2 (final temperature) = -5°C = -5 + 273 = 268 K
V2 (final volume) = ?
2. Use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Plugging in the given values:
(755 torr * 58.7 L) / 296 K = (0.051 atm * V2) / 268 K
Rearranging the equation to solve for V2:
V2 = (755 torr * 58.7 L * 268 K) / (296 K * 0.051 atm)
Calculating the value:
V2 ≈ (10786156.44 torr * L * K) / (1512.296 atm) ≈ 7136.8 L
Therefore, the balloon will reach a volume of approximately 7136.8 L, which is greater than its maximum volume of 788 L. So, it will exceed its maximum volume when it reaches the given altitude.
I hope this helps! Let me know if you have any further questions.