Question

The gauge pressure in your car tires is 2.70 ✕ 105 N/m2 at a temperature of 35.0°C when you drive it onto a ship in Los Angeles to be sent to Alaska. What is their gauge pressure (in atm) later, when their temperature has dropped to −42.0°C? Assume the tires have not gained or lost any air.

Answer 1

The final gauge pressure will be 1.74 atm

Step-by-step explanation:

Assume air tire as an ideal gas, therefore, it is considered:

• incompressible gas: the gas density will be constant.

• there is no change in the amount of matter: the number of moles is constant (because there is no gain or loss of air during the process)

The expression to describe the ideal gas process is:

`P.v=Rg.T`

Where v is the specific volume or the inverse of the density p:

`P/p=Rg.T` (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:

`P/T=Rg.p`

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:

`P_(1)/T_(1)=P_(2)/T_(2)` (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:

`T(K)=273+T(°C)`

`T_(1) (K)=273+35`

`T_(1)=308 K`

`T_(2) (K)=273-42`

`T_(2)=231 K`

Note also that the pressure given is the gauge pressure therefore it must be expressed as absolute pressure:

`Pa (Pa)=Patm(Pa)+Pg(Pa)`

Where Patm is the atmosphere pressure and is equal to 101325 Pa then for Pg of 2.7 10^5 Pa:

`Pa_(1) (Pa)= 101325 Pa+ 270000 Pa`

`Pa_(1) (Pa)= 371325 Pa`

Solving equation (3) for pressure at state 2:

`P_(2)=P_(1)*T_(2)/T_(1)`

For the temperatures and pressure calculated values:

`P_(2)= 371325 Pa * \frac231K}/{308K}`

`P_(2)= 278493.75 Pa`

As is required the gauge pressure and not the absolute pressure:

`Pg (Pa)=Pa(Pa)-Patm(Pa)`

`Pg_(2) (Pa)= 278493.75 -101325(Pa)`

`Pg_(2) (Pa)= 177168 Pa`

Finally re expressing the pressure in atm units:

`P (atm)=P (Pa)*(1 atm)/(101325 Pa)`

`P_(2) (atm)=177168*(1 atm)/(101325 Pa)`

`P_(2) (atm)=1.74 atm`