Question

A normal breath has a volume of about 1 L. The pressure exerted by the lungs to draw air in is about 758 torr. Assuming that the outside air pressure is 760 torr, calculate the change in entropy of a breath of air when it is inhaled into the lungs. Assume that the air remains at a temperature of ${25}^{0}C$ and that it behaves ideally.

• A. 0.0009 J/K
• B. 0.0008 J/K
• C. 0.0010 J/K
• D. None of the above

Answer 1

The correct option is A 0.0009 J/K

Assuming that the gas behaves ideally and that the process is reversible, we have that the entropy change in an isothermal reversible process is

$\Delta S=nR\ln \frac{V_2}{V_1}$

If the process is isothermal then we can use the ideal gas equation to get

$\Delta S=nR\ln \frac{P_2}{P_1}$

Now in order to use the formula, we need to determine the moles in 1 L of air at 25°C and an initial pressure of 760 torr (=1 atm). Using the ideal gas law,

$n=\frac{PV}{RT}=\frac{1\times 1}{0.08205\times 298}=0.0409mol$

Substituting in entropy formula,

$\Delta S=nR\ln \frac{P_2}{P_1}=0.0409\times 8.314\times \ln \frac{760}{758}=0.0009J{K}^{-1}$