Answer:
1. Using the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin, we can solve for V:
V = (nRT) / P
Plugging in the given values, we get:
V = (0.210 mol)(0.08206 L·atm/mol·K)(295.5 K) / 1.00 atm
V = 4.91 L
Therefore, the sample of gas occupies 4.91 L.
2. We can use the combined gas law to find the new volume of the sample at the new temperature and pressure:
(P1V1) / (n1T1) = (P2V2) / (n2T2)
At conditions 1, we know P1 = 1.00 atm, V1 = 4.91 L, n1 = 0.210 mol, and T1 = 295.5 K.
At conditions 2, we know P2 = 1.00 atm (pressure is constant), n2 = n1 (number of moles is constant), and T2 = 95.0 + 273.15 = 368.15 K.
Solving for V2, we get:
V2 = (P1V1n2T2) / (n1T1P2)
V2 = (1.00 atm)(4.91 L)(0.210 mol)(368.15 K) / (0.210 mol)(295.5 K)(1.00 atm)
V2 = 7.23 L
Therefore, the new volume of the sample at the higher temperature is 7.23 L.
3. At STP (standard temperature and pressure), which is 0 °C (273.15 K) and 1 atm, the molar volume of an ideal gas is 22.4 L/mol. Therefore, the number of moles of gas can be calculated as:
n = V / Vm
where Vm is the molar volume of gas at STP, which is 22.4 L/mol.
n = 2.82 L / 22.4 L/mol
n = 0.126 mol
The molar mass of the gas can be calculated as:
molar mass = mass / n
mass = 2.54 g
molar mass = 2.54 g / 0.126 mol
molar mass = 20.16 g/mol
Therefore, the molar mass of the gas is 20.16 g/mol.
4. Let's assume that the total number of moles in the mixture is 1 (since we don't know the actual amount). The first gas constitutes 65% of the mixture, so its mole fraction is 0.65. Therefore, the mole fraction of the second gas is 0.35 (since the sum of the mole fractions must equal 1). The partial pressures of the two gases can be calculated using the mole fractions and the total pressure:
Pgas1 = mole fraction of gas 1 x total pressure
Pgas1 = 0.65 x 4.3 atm
Pgas1 = 2.79 atm
Pgas2 = mole fraction of gas 2 x total pressure
Pgas2 = 0.35 x 4.3 atm
Pgas2 = 1.51 atm
Therefore, the partial pressure of the first gas is 2.79 atm and the partial pressure of the second gas is 1.51 atm.
5. The mole fraction of a gas is the ratio of the number of moles of that gas to the total number of moles in the mixture. We can find the total number of moles by adding up the partial pressures of the gases and dividing by the total pressure:
total moles = (partial pressure of gas A + partial pressure of gas B + partial pressure of gas C) / total pressure
total moles = (0.75 atm + 0.50 atm + 0.25 atm) / 1.50 atm
total moles = 1.50 moles
To find the mole fraction of helium, we need to know the number of moles of helium. We can use the partial pressure of helium and the total pressure to calculate the number of moles of helium using the ideal gas law:
n = PV / RT
n = (0.50 atm)(1.00 L) / (0.08206 L·atm/mol·K)(298 K)
n = 0.0202 mol
Now we can calculate the mole fraction of helium:
mole fraction of helium = moles of helium / total moles
mole fraction of helium = 0.0202 mol / 1.50 moles
mole fraction of helium = 0.0135
Therefore, the mole fraction of helium in the gas mixture is 0.0135.