A car tire contains 0.0380 m^3 of air at a pressure of 2.20 ×10^5 Pa (about 32 psi). How much mo

step 1 Given that in this problem we have given V1 is equal to 0 .0380 meter cube, V2 is equal to V1 an

A car tire contains 0.0380 m^3 of air at a pressure of 2.20...

A car tire contains $0.0380 \mathrm{m}^{3}$ of air at a pressure of $2.20 \times 10^{5} \mathrm{Pa}$ (about 32 psi). How much more internal energy does this gas have than the same volume has at zero gauge pressure (which is equivalent to normal atmospheric pressure)?

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Ideal Gas Law

The ideal gas law, given by PV = nRT, relates the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas. In problems like this, it is used to calculate the temperature change or the number of moles of gas when pressure or volume conditions change, allowing one to assess how the state of the gas is altered between two conditions.

Internal Energy of an Ideal Gas

For an ideal gas, the internal energy is a function only of its temperature and the degrees of freedom of its molecules. It is typically expressed as U = (f/2)nRT, where f is the degrees of freedom. This relationship is crucial when comparing different states of the gas, as any increase in internal energy reflects a corresponding increase in temperature despite the gas being contained in a constant volume.

Gauge vs. Absolute Pressure

Gauge pressure is the pressure measured relative to atmospheric pressure. In contrast, absolute pressure includes atmospheric pressure in its measurement. Recognizing this distinction is important when dealing with problems that compare a scenario with a given gauge pressure to one at atmospheric conditions, as it directly affects the computed state variables such as temperature and, consequently, the internal energy.

Transcript

Please give Ace some feedback