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.

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!

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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.

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