Question

1. a car tire contains ( 0.0380 mathrm{~m}^{3} ) of air at a pressure of ( 2.20 \times 10^{5} mathrm{~n} / mathrm{m}^{2} ) (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)?

Explanation:

Step1: Calculate the number of moles using the ideal gas law

The ideal gas law is \(PV = nRT\). We assume the temperature \(T\) is constant. Let \(P_1 = 2.20\times 10^{5}\space N/m^{2}\), \(V = 0.0380\space m^{3}\), and \(P_2= 1.01\times 10^{5}\space N/m^{2}\) (atmospheric pressure).

For the first state \(n_1=\frac{P_1V}{RT}\), for the second state \(n_2=\frac{P_2V}{RT}\)

Step2: Use the formula for internal energy of an ideal gas

The internal energy of an ideal gas is \(U=\frac{5}{2}nRT\) (for diatomic gas, like air).

The internal energy difference \(\Delta U=\frac{5}{2}(n_1 - n_2)RT\)

Substituting \(n_1\) and \(n_2\) from step 1: \(\Delta U=\frac{5}{2}\frac{(P_1 - P_2)V}{RT}\times RT=\frac{5}{2}(P_1 - P_2)V\)

Substitute \(P_1 = 2.20\times 10^{5}\space N/m^{2}\), \(P_2= 1.01\times 10^{5}\space N/m^{2}\), \(V = 0.0380\space m^{3}\)

\(\Delta U=\frac{5}{2}(2.20\times 10^{5}- 1.01\times 10^{5})\times0.0380\)

\(\Delta U=\frac{5}{2}\times1.19\times 10^{5}\times0.0380\)

\(\Delta U=\frac{5\times1.19\times 10^{5}\times0.0380}{2}\)

\(\Delta U=\frac{5\times4522}{2}\)

\(\Delta U = 11305\space J\)

Answer:

\(11305\space J\)