T1 = 27 + 273 = 300 K, T2 = 127 + 273 = 400 K[1]
Apply V1T1 = V2T2
V2 = 2.00 × 400300
= 2.67 dm3[1]
T = 23 + 273 = 296 K, V = 246 × 10−6 m3
Using pV = nRT = mMrRT,
Mr = 0.320 × 8.31 × 2961.00 × 105 × 246 × 10−6
= 32.0[1]
The gas is O2.[1]
Apply p1V1T1 = p2V2T2, with T1 = 300 K and T2 = 400 K[1]
p2 = 100 × 2.000.500 × 400300
= 533 kPa[1]
pO2 = mole fraction of O2 × total pressure
= 0.21 × 101
= 21.2 kPa[1]
NH3 deviates more.
B87.5 kPa[1]
Using p1V1 = p2V2,
50 × 10 = psmall(10 + 30), so psmall = 50040 = 12.5 kPa
100 × 30 = plarge(10 + 30), so plarge = 300040 = 75 kPa
ptotal = 12.5 + 75 = 87.5 kPa
C1 g of H2 and 2 g of D2 in a total volume V[1]
Pressure exerted by 2 g of H2, p = (2 ÷ 2(1.0))RTV = RTV
Amount of oxygen = 12016.0 × 2 = 3.75 mol
Amount of hydrogen = 5.001.0 × 2 = 2.50 mol
Mole fraction of oxygen, xO2 = 3.753.75 + 2.50 = 0.600[1]
pO2 = xO2 × pT
3.00 = 0.600 × pT
pT = 5.00 atm[1]
2H2(g) + O2(g) → 2H2O(l)
H2 is limiting and is used up.
Amount of oxygen remaining = 3.75 − 2.502 = 2.50 mol[1]
The water condenses on cooling, so the final pressure is due to 2.50 mol of O2 only.
Pressure = 2.503.75 × 3.00 = 2.00 atm[1]
Low pressure and high temperature[1]
At low pressure, the gas molecules are far apart. The volume of the gas molecules therefore becomes insignificant compared to the volume of the container or the intermolecular forces of attraction are negligible.[1]
At high temperature, the gas molecules move faster (higher kinetic energy) and the intermolecular forces of attraction become negligible.[1]
Using the ideal gas equation,
p = nRTV = 1.00 × 8.31 × 2980.250 × 10−3
= 9.91 × 106 Pa[1]
| C | H | |
|---|---|---|
| % by mass | 85.7 | 14.3 |
| Ar | 12.0 | 1.0 |
| no. of moles | 7.142 | 14.3 |
| simplest ratio | 1 | 2 |
Empirical formula of A: CH2[1]
CxHy + (x + y4)O2 → xCO2 + y2H2O
Volume of CO2 produced = 65.0 − 35.0 = 30.0 cm3
Volume of O2 reacted = 80.0 − 35.0 = 45.0 cm3[1]
nAnCO2 = 10.030.0 = 1x, so x = 3
nAnO2 = 10.045.0 = 13 + y/4, so y = 6
Molecular formula of A: C3H6[1]
Increasing deviation: He < Ne < O2 < C2H6 < NH3 < H2O[1]
H2O and NH3 deviate more than the rest: they have stronger hydrogen bonding, compared with the weaker instantaneous dipole–induced dipole interactions in the others. H2O has more extensive and stronger hydrogen bonding than NH3.[1]
He < Ne < O2 < C2H6: as the number of electrons increases, the electron cloud is larger and more polarisable, so the instantaneous dipole–induced dipole interactions are stronger.[1]
For an ideal gas, pV stays constant as p increases.
Ne: as pressure increases (volume decreases), the volume of the Ne atoms becomes significant compared with the total volume of the container. The pressure exerted is greater than for an ideal gas, so pV is greater than ideal.[1]
O2 and NH3: as pressure increases, the molecules are pushed closer together, so the intermolecular forces of attraction become significant. The molecules collide with the walls less forcefully, so the pressure is lower than ideal and pV decreases.[1]
NH3 deviates more than O2 because of the stronger hydrogen bonding between NH3 molecules, compared with the weaker instantaneous dipole–induced dipole interactions between O2 molecules.[1]
NoteNe has only very weak instantaneous dipole–induced dipole interactions (a very small electron cloud), so the volume of its atoms has the larger effect.
pV = nRT
(6.00 × 105)(670 × 10−6) = n(8.31)(273 + 20)[1]
n = 0.165 mol[1]
p1T1 = p2T2
6.0020 + 273 = p at 10 000 m5 + 273[1]
p at 10 000 m = 5.69 bar = 5.69 × 105 Pa[1]
It is not necessary to deflate the bicycle tyre.
The maximum difference between the internal and external pressure the tyre can endure without bursting = 8 − 1.01 = 6.99 bar[1]
At 10 000 m, the difference between the internal and external pressure of the tyre = 5.693 − 0.28 = 5.41 bar, which is within what the tyre can endure without bursting.[1]
Exam tipThe tyre bursts because of the pressure difference, so compare differences, not the internal pressures alone.