The gauge pressure in your car tires is 3.00 ✕ 10^5 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.

Explanation:

Using the ideal gas equation, which I presume you are since you didn't specify using any other EOS, we have PV=nRT. Solving for what changes, i.e. pressure(P) and temperature(T), we have P/T=nR/V. Now, we can set up a relationship between the two pressures and temperatures and solve for what's necessary.

So, we have:

P1/T1=P2/T2

Solving for P2, we have:

P2=(P1*T2)/T1

NOTE: We MUST convert our temperatures to kelvin, otherwise you will end up with a NEGATIVE AND INCORRECT pressure!

Plugging in our values of P1=3.00x10^5 N/m^2, T1 of 308.15K, and T2 of 235.15K. Now we are free to evaluate:

P2=[(3.00x10*5 N/m^2)(235.15K)]/[308.15K]

P2=228930.7156 N/m^2

Or, to the appropriate amount of significant figures: 2.29x10^5 N/m^2

Which makes sense intuitively, as things tend to deflate slightly when the temperature drops!

Hope this helps!

Mark as brainlist...

the ima (higher than 1m) will be greater than the ima with the height of 1m.