Answer:
The final gauge pressure will be 1.74 atm
Explanation:
Assume air tire as an ideal gas, therefore, it is considered:
The expression to describe the ideal gas process is:
[tex]P.v=Rg.T[/tex]
Where v is the specific volume or the inverse of the density p:
[tex]P/p=Rg.T[/tex] (2)
Then the P and T are the absolute pressure and temperature respectively. Rg represents the particular gas constant for air, Rg is equal to 287 J/kg-K. Take into account that P and T must be expressed in Pascal and Kelvin respectively.
By reorganizing the expression (2) as below is doing:
[tex]P/T=Rg.p[/tex]
Can be noticed that the product Rg*p is constant therefore relation P/T will also be constant. For two different states 1 and 2 of the ideal gas, it follows:
[tex]P_{1}/T_{1}=P_{2}/T_{2}[/tex] (3)
Here, states 1 and 2 will represent the state before and after arriving in Alaska.
Note that the temperatures given are in °C so it must be converted:
[tex]T(K)=273+T(°C)[/tex]
[tex]T_{1} (K)=273+35[/tex]
[tex]T_{1}=308 K[/tex]
[tex]T_{2} (K)=273-42[/tex]
[tex]T_{2}=231 K[/tex]
Note also that the pressure given is the gauge pressure therefore it must be expressed as absolute pressure:
[tex]Pa (Pa)=Patm(Pa)+Pg(Pa)[/tex]
Where Patm is the atmosphere pressure and is equal to 101325 Pa then for Pg of 2.7 10^5 Pa:
[tex]Pa_{1} (Pa)= 101325 Pa+ 270000 Pa[/tex]
[tex]Pa_{1} (Pa)= 371325 Pa[/tex]
Solving equation (3) for pressure at state 2:
[tex]P_{2}=P_{1}*T_{2}/T_{1}[/tex]
For the temperatures and pressure calculated values:
[tex]P_{2}= 371325 Pa * \frac231K}/{308K}[/tex]
[tex]P_{2}= 278493.75 Pa [/tex]
As is required the gauge pressure and not the absolute pressure:
[tex]Pg (Pa)=Pa(Pa)-Patm(Pa)[/tex]
[tex]Pg_{2} (Pa)= 278493.75 -101325(Pa)[/tex]
[tex]Pg_{2} (Pa)= 177168 Pa[/tex]
Finally re expressing the pressure in atm units:
[tex]P (atm)=P (Pa)*\frac{1 atm}{101325 Pa}[/tex]
[tex]P_{2} (atm)=177168*\frac{1 atm}{101325 Pa}[/tex]
[tex]P_{2} (atm)=1.74 atm[/tex]
