D800 kPa[1]
At constant T, apply p1V1 = p2V2 to each gas.
pO2 = 200 × 5.02.5 = 400 kPa
pN2 = 500 × 2.02.5 = 400 kPa
Total pressure = 400 + 400 = 800 kPa
CNO2[1]
pV = nRT, where V, R and T are the same for all four gases, so p ∝ n.
The greatest Mr gives the smallest number of moles for the same mass, and hence the lowest pressure.
| gas | H2 | CH4 | NO2 | Ar |
|---|---|---|---|---|
| Mr or Ar | 2.0 | 16.0 | 46.0 | 39.9 |
| n = 5.00 ÷ Mr | 2.50 | 0.313 | 0.109 | 0.125 |
Bhelium[1]
An ideal gas should have negligible intermolecular forces of attraction.
Ammonia has hydrogen bonding, which has the strongest intermolecular forces of attraction.
Between helium, krypton and nitrogen, helium has the smallest electron cloud, so the weakest instantaneous dipole–induced dipole interactions.
V = (5.2)(8.31)(15 + 273)101 300 = 0.123 m3
= 123 dm3[1]
Molecules of an ideal gas have zero (negligible) volume.[1]
There are negligible (no) intermolecular forces of attraction between ideal gas molecules.[1]
Collisions between ideal gas molecules are perfectly elastic (no loss of energy during collision).[1]
Hydrogen bond drawn from a lone pair on F of one molecule to the H of the other, and labelled.[1]
Lone pair and dipoles (δ+ on H, δ− on F) shown.[1]
Correct sketch: HF dips further below 1.0 than fluorine.[1]
HF deviates more from ideal behaviour because it has stronger intermolecular hydrogen bonding between its molecules, compared to F2, which has only instantaneous dipole–induced dipole interactions between its molecules.[1]
Horizontal line at pV/T = nR = 8.31 J K−1 for 1 mol.[1]
NH3 graph below the ideal line at moderate pressure.[1]
At moderately high pressure, NH3 molecules come closer together and the intermolecular attractive forces (hydrogen bonds) between NH3 molecules become significant. This causes the gas to occupy a volume smaller than that of an ideal gas.[1]
27 °C = 300 K and 227 °C = 500 K
When compressed at constant T to half V, apply p1V1 = p2V2
4.00 × 5.00 = p2 × 2.50
p2 = 8.00 atm[1]
When heated from 300 K to 500 K at constant V, apply p2T2 = p3T3
8.00300 = p3500
p3 = 13.3 atm[1]
Volume of O2 consumed per minute = 21 − 14100 × 500 × 12
= 420 cm3 = 4.20 × 10−4 m3[1]
pV = nRT
(101 325)(4.20 × 10−4) = n(8.31)(37 + 273)[1]
n = 0.0165 mol (1.652 × 10−2 mol)[1]
From reaction 1, 6 mol of O2 releases 2818 kJ of energy.
Energy released per minute = 28186 × 0.01652
= 7.76 kJ[1]
Correct sketch: bromobutane dips below 1.0, but less than iodobutane.[1]
Any two of:[2]
There are significant intermolecular forces of attraction (hydrogen bonding) between molecules, which are not negligible.
orThe particles of N2H4 take up space relative to the volume of the gas, which is not negligible.
orCollisions are inelastic.
Down the group, the atomic size increases. Hence, the volume of the gas particles becomes less negligible (more significant) compared to the volume of the container.[1]
Also, down the group, the number of electrons in the atoms increases and the electron cloud becomes bigger and more polarisable, leading to stronger instantaneous dipole–induced dipole interactions. Hence, the intermolecular forces of attraction become less negligible (more significant).[1]
n = pVRT = 8.00 × 104 × 1.00 × 10−38.31 × 300 = 0.03209 mol[1]
Mr = 2.790.03209 = 86.9[1]
Ar = 80(2.4) + 82(11.7) + 83(11.6) + 84(56.9) + 86(17.4)100
= 83.9[1]
n = 1.00 × 105 × 1.00 × 10−38.31 × 293 = 0.04107 mol[1]
Ar = 3.400.04107 = 82.8[1]
The sample is not likely to have the same isotopic composition as that in (b), as its average Ar is different from the average Ar calculated in (b).[1]
Since Xe and F2 are mixed in equal volumes, the partial pressure of both gases is the same. Let it be p.
Initial total pressure = p + p = 2p
Final total pressure = 0.70 × 2p = 1.4p
Final partial pressure of Xe = 47(1.4p) = 0.8p, final partial pressure of F2 = 1.4p − 0.8p = 0.6p[1]
V and T are constant, so pV = nRT gives p ∝ n.
Amount of Xe reacted ∝ p − 0.8p = 0.2p
Amount of F2 reacted ∝ p − 0.6p = 0.4p, so 1 mol of Xe reacts with 2 mol of F2.[1]
Xe(g) + 2F2(g) → XeF4(s)[1]
Exam tipAt constant V and T, pressure is proportional to amount, so partial pressures can be used in place of moles.