Question 0a30f

Mar 1, 2017

(a) The pressure rise in the tyre is 149 kPa. (b) The mass of air that must be bled off is 17 g.

Explanation:

(a) Calculate the pressure rise.

Since the volume is constant but the pressure and temperature are changing, this is an example of Gay-Lussac's Law:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {P}_{1} / {T}_{1} = {P}_{2} / {T}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange the above formula to get

P_2=P_1 × T_2/T_1

A gauge pressure (${P}_{\text{g}}$) is the pressure difference between a system ${P}_{\text{s}}$ and the surrounding atmosphere ${P}_{\text{atm}}$.

${P}_{\text{g" = P_"s" - P_"atm}}$

or

${P}_{\text{s" = P_"g" + P_"atm}}$

${P}_{1} = \text{(210 + 101.3) kPa = 311.3 kPa}$
${T}_{1} = \text{(250 + 273.15) K" = "523.15 K}$
${P}_{2} = \text{?}$
${T}_{2} = \text{(500 + 273.15) K" = "773.15 K}$

${P}_{2} = \text{311.3 kPa" × (773.15 color(red)(cancel(color(black)("K"))))/(523.15 color(red)(cancel(color(black)("K")))) = "460.1 kPa}$

The new pressure is 460.1 kPa.

The pressure rise is

ΔP = P_2 - P_1 = "(460.1 - 311.3) kPa = 149 kPa"

(b) Calculate the mass of air to be bled off.

For this calculation, we can use the Ideal Gas Law:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Since

$\text{moles" = "mass"/"molar mass}$ or $n = \frac{m}{M}$,

We can re-write the Ideal Gas Law as

$P V = \frac{m}{M} R T$

or

$m = \frac{P V M}{R T}$

Since V, M, R" and $T$ are constant, we can write

Δm = ΔP × (VM)/(RT)

Δm = 149 × 10^3 color(red)(cancel(color(black)("Pa"))) × (0.025 color(red)(cancel(color(black)("m"^3))) × "29 g"·color(red)(cancel(color(black)("mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("Pa·m"^3"K"^"-1""mol"^"-1"))) × 773.15 color(red)(cancel(color(black)("K")))) = "17 g"#

You would have to bleed off 17 g of air.