Question

A car tire contains 0.0380m3 of air at a pressure of 2.20×105N/m2 (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)?

Answer 1

Answer:

$\Delta U = 11305 J$

Explanation:

As per ideal gas equation we know that

PV = nRT

here we know that

$P = 2.20 \times 10^5 Pa$

$V = 0.0380 m^3$

now we have

$nRT_1 = (2.20 \times 10^5)(0.0380) = 8360 J$

now for other case at zero gauge pressure we have

$P = 1.01 \times 10^5 Pa$

$V = 0.0380 m^3$

we have now

$nRT_2 = (1.01 \times 10^5)(0.0380) = 3838 J$

now the difference in internal energy of air is given as

$\Delta U = \frac{5}{2}nR\Delta T$

$\Delta U = \frac{5}{2}(8360 - 3838)$

$\Delta U = 11305 J$