IB Chemistry HL 1.5 Ideal Gases Questions

Practise IB Chemistry HL 1.5 with evidence-led questions on the ideal gas equation, molar volume and deviations from ideal behaviour.

Exam points

Use PV=nRT with consistent units to calculate gas variables, amount or molar mass.
Relate gas volume and amount at stated conditions, including molar-volume calculations.
Explain when gases deviate from ideal behaviour using pressure, temperature, molecular volume and attractions.

Covered syllabus points

1.5.1—Ideal gas model assumptions 1.5.2—Real gases 1.5.3—Molar volume 1.5.4—Ideal gas equation

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Study 1.5 Ideal gases conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

[Maximum number: 4]

Alkanes are commonly occurring organic compounds.

(a)

The first four straight chain alkanes are gases at room temperature.

Boiling points of straight chain alkanes

[ 4 ]

(i)

Calculate the volume, in $\mathrm{dm}^{3}$ , occupied by 6.45 g of propane gas at 100 kPa and $15^{\circ} \mathrm{C}$ .

[ 2 ]

$\begin{aligned} \text { « } \mathrm{Mr}\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)=44.11 \text { » } \text { « } \mathrm{n}=\text { » } 0.146 \text { «mol» } \text { «V }=\frac{0.146 \times 8.31 \times 288}{100}=\text { » } 3.49 \text { «dm³» } \end{aligned}$

Award [2] for correct final answer.

Accept answers in the range 3.49-3.59 «dm³».

(ii)

Outline why the volume occupied by propane(g) at very high pressure is higher than the value calculated using PV=nRT.

[ 2 ]

[Maximum number: 1]

Two hydrides of nitrogen are ammonia and hydrazine, $\mathrm{N}_{2} \mathrm{H}_{4}$ . One derivative of ammonia is methanamine whose molecular structure is shown below.

Calculate, showing your working, the mass of hydrazine needed to remove all the dissolved oxygen from $1000 \mathrm{dm}^{3}$ of the sample.

Calculate the volume, in $\mathrm{dm}^{3}$ , of nitrogen formed under SATP conditions.
(The volume of 1 mol of gas $=24.8 \mathrm{dm}^{3}$ at SATP.)

$n(\mathrm{N}_{2}\mathrm{H}_{4})=n(\mathrm{O}_{2}) =\frac{8.0\ \mathrm{g}}{32.00\ \mathrm{g\ mol^{-1}}} =0.25\ \mathrm{mol}$ $\text{volume of nitrogen} =0.25\ \mathrm{mol}\times24.8\ \mathrm{dm^{3}\ mol^{-1}} =6.2\ \mathrm{dm^{3}}$

[Maximum number: 3]

A group of students investigated the rate of the reaction between aqueous sodium thiosulfate and hydrochloric acid according to the equation below.

\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(\mathrm{aq})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow 2 \mathrm{NaCl}(\mathrm{aq})+\mathrm{SO}_{2}(\mathrm{~g})+\mathrm{S}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(1)

The two reagents were rapidly mixed together in a beaker and placed over a mark on a piece of paper. The time taken for the precipitate of sulfur to obscure the mark when viewed through the reaction mixture was recorded.

Initially they measured out $10.0 \mathrm{~cm}^{3}$ of $0.500 \mathrm{moldm}^{-3}$ hydrochloric acid and then added $40.0 \mathrm{~cm}^{3}$ of $0.0200 \mathrm{~mol} \mathrm{dm} \mathrm{m}^{-3}$ aqueous sodium thiosulfate. The mark on the paper was obscured 47 seconds after the solutions were mixed.

Another group suggested collecting the sulfur dioxide and drawing a graph of the volume of gas against time.

Calculate the volume of sulfur dioxide, in $\mathrm{cm}^{3}$ , that the original reaction mixture would produce if it were collected at $1.00 \times 10^{5} \mathrm{~Pa}$ and 300 K .

Calculate the amount of sulfur dioxide:

$n(\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}) =n(\mathrm{SO}_{2}) =0.0400\times0.0200 =0.000800\ \mathrm{mol}$ $V=\frac{nRT}{P} =\frac{0.000800\times8.31\times300}{10^{5}} =1.99\times10^{-5}\ \mathrm{m}^{3} =19.9\ \mathrm{cm}^{3}$

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