1) 1.296 mol
2) [tex]2.69\cdot 10^5 Pa[/tex]
3) Yes
4) 234 kPa
Explanation:
1)
In order to find the number of moles inside each tire, we can use the equation of state for an ideal gas:
[tex]pV=nRT[/tex]
where:
p is the pressure of the gas
V its volume
n the number of moles
R the gas constant
T the absolute temperature
At the beginning, we have:
[tex]p=240 kPa = 2.40\cdot 10^5 Pa[/tex]
[tex]V=13.2 L = 0.0132 m^2[/tex]
[tex]R=8.314 J/mol K[/tex]
[tex]T=21^{\circ}C+273=294 K[/tex]
Therefore, the number of moles of nitrogen in each tire is:
[tex]n=\frac{pV}{RT}=\frac{(2.40\cdot 10^5)(0.0132)}{(8.314)(294)}=1.296mol[/tex]
2)
The equation of state can be rewritten as
[tex]\frac{p}{T}=\frac{nR}{V}[/tex]
For the nitrogen gas inside the tires, the quantity nR/V remains constant, so we can write:
[tex]\frac{p_1}{T_1}=\frac{p_2}{T_2}[/tex]
Where in this problem:
[tex]p_1 = 2.40\cdot 10^5 Pa[/tex] is the initial pressure
[tex]T_1=294 K[/tex] is the initial temperature
[tex]p_2[/tex] is the final pressure in Death Valley
[tex]T_2=56^{\circ}C+273=329 K[/tex] is the temperature in Death Valley
Solving for p2, we find the final pressure of the tires:
[tex]p_2 = \frac{p_1 T_2}{T_1}=\frac{(2.40\cdot 10^5)(329)}{294}=2.69\cdot 10^5 Pa[/tex]
3)
As we have calculated in part 2, the pressure of the gas inside the tires when the car reaches the Death Valley will be
[tex]p_2 = 2.69\cdot 10^5 Pa[/tex]
Which can be rewritten as
[tex]p_2 = 269 kPa[/tex]
The text of the problem states that the tires will burst if the internal pressure reaches a value of
[tex]p_b=262 kPa[/tex]
We observe that
[tex]p_2>p_b[/tex]
Which means that the internal pressure is larger than the breaking pressure: so, the tires will burst.
4)
Here we want the final pressure in the tires to be at most equal to the breaking pressure: so it must be
[tex]p_2 = p_b = 262 kPa[/tex]
We can use again the equation used in part 2:
[tex]\frac{p_1}{T_1}=\frac{p_2}{T_2}[/tex]
In order to find [tex]p_1[/tex], the initial pressure at which the tires should be in order not to burst when the car arrives in the Death Valley.
The temperatures are
[tex]T_1=294 K[/tex] is the initial temperature
[tex]T_2=56^{\circ}C+273=329 K[/tex] is the temperature in Death Valley
So the initial pressure must be
[tex]p_1 = \frac{p_2 T_1}{T_2}=\frac{(262)(294)}{329}=234 kPa[/tex]
