The correct answer is A because:

The equation for the kinetic energy of a gas molecule is
- $\frac{1}{2}$ m ${(C_{r m s})}^{2}$ = $\frac{3}{2}$ kT
Rearranging this equation, we get
- ${(C_{r m s})}^{2} = \frac{3 k T}{m}$
We are told that there is no air lost from the tyre, therefore m will be constant before and after the journey
To use the above equation, we need to convert the temperature to Kelvin
This is done by adding 273 to the °C temperature, hence
- 10°C = 283 K
- 30°C = 303 K
The ratio of the final ${(C_{r m s})}^{2}$ to the initial ${(C_{r m s})}^{2}$ will be
- $\frac{f i n a l {(c_{r m s})}^{2}}{i n i t i a l {(c_{r m s})}^{2}}$ $= \frac{T_{f i n a l}}{T_{i n i t i a l}}$
- $\frac{f i n a l {(c_{r m s})}^{2}}{i n i t i a l {(c_{r m s})}^{2}}$ = $\frac{303}{283}$ = 1.07
- Hence, ${(c_{r m s})}_{f i n a l}^{2}$ = 1.07 ${(c_{r m s})}_{i n i t i a l}^{2}$
The root–mean–square speed of the molecules is is the square root of ${(C_{r m s})}^{2}$ , therefore
- ${(c_{r m s})}_{f i n a l} = \sqrt{1.07} {(c_{r m s})}_{i n i t i a l}$
The percentage increase in the $C_{r m s}$ of the air molecules can be calculated from
- Δ $C_{r m s}$ % = $\frac{{(c_{r m s})}_{f i n a l} - {(c_{r m s})}_{i n i t i a l}}{{(c_{r m s})}_{i n i t i a l}}$ × 100
- Δ $C_{r m s}$ % = $\frac{\sqrt{1.07} - 1}{1}$ × 100

Δ $C_{r m s}$ % = 3.47 %

Molecular Kinetic Theory Model (AQA A Level Physics)

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