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When air is inhaled, it enters the alveoli of the lungs, and varying amounts of the component gases exchange with dissolved gases in the blood. The resulting alveolar gas mixture is quite different from the atmospheric mixture. The following table presents selected data on the composition and partial pressure of four gases in the atmosphere and in the alveoli:

 Gas  Mole %  Partial  Pressure (torr)  Mole %  Partial  Pressure (torr) N278.6569O220.9104CO200.0440H2O00.4647\begin{array}{|c|c|c|c|c|} \hline \text { Gas } & \text { Mole \% } & \begin{array}{l} \text { Partial } \\ \text { Pressure (torr) } \end{array} & \text { Mole \% } & \begin{array}{l} \text { Partial } \\ \text { Pressure (torr) } \end{array} \\ \hline \mathrm{N}_2 & 78.6 & - & - & 569 \\ \hline \mathrm{O}_2 & 20.9 & - & - & 104 \\ \hline \mathrm{CO}_2 & 00.04 & - & - & 40 \\ \hline \mathrm{H}_2 \mathrm{O} & 00.46 & - & - & 47 \\ \hline \end{array}

If the total pressure of each gas mixture is 1.00 atm1.00 \mathrm{~atm}, calculate:

(a) The partial pressure (in torr) of each gas in the atmosphere

(b) The mole % of each gas in the alveoli

(c) The number of O2\mathrm{O}_2 molecules in 0.50 L0.50 \mathrm{~L} of alveolar air (volume of an average breath of a person at rest) at 37C37^{\circ} \mathrm{C}

Question

The cycle in the given figure represents the operation of a gasoline internal combustion engine. Assume the gasoline air intake mixture is an ideal gas with y=1.30. What are the ratios of What is the engine efficiency?

Solution

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f) Since the efficiency is the ratio of work done by the gas and the heat added into the system, we need to find those two quantities. The total work is the sum of the work done from 22 to 33 and 44 to 11. Processes from 11 to 22 and 33 to 44 are isochoric, meaning the volume remains constant, and the work done is then zero. Using the third equation, the relationships pVγ=p2V2γpV^{\gamma} = p_2V_2^{\gamma} as well pVγ=p4V2γpV^{\gamma} = p_4V_2^{\gamma} must be true. Solving both equations for pp and plugging them into the last equation we have:

W=W2,3+W4,1=V2V3p2V2γVγdV+V4V1p4V4γVγdV=p2V2γV2V3VγdV+p4V4γV4V1VγdV=p2V2γ(1γ1V1γV2V3)+p4V4γ(1γ1V1γV4V1)=(p2V2γγ1)(V21γV31γ)+(p4V4γγ1)(V41γV11γ)=(3p1V11γ)(114γ1)(p1V1γ1)(114γ1)=(3nRT1γ1)(114γ1)(nRT1γ1)(114γ1)=(2nRT1γ1)(114γ1)\begin{align*} W &= W_{2,3} + W_{4,1} \\ &= \int\limits_{V_2}^{ V_3} {\frac{ p_2V_2^{\gamma}}{ V^{\gamma}} \text d V} + \int\limits_{V_4}^{ V_1} {\frac{ p_4V_4^{\gamma}}{ V^{\gamma}} \text d V} \\ &= p_2V_2^{\gamma} \int\limits_{V_2}^{ V_3} { V^{- \gamma} \text d V} + p_4V_4^{\gamma} \int\limits_{V_4}^{ V_1} { V^{- \gamma} \text d V} \\ &= p_2V_2^{\gamma} \left( \frac{1}{\gamma - 1} V^{1- \gamma} \Big|_{ V_2}^{ V_3} \right) + p_4V_4^{\gamma} \left( \frac{1}{\gamma - 1} V^{1- \gamma} \Big|_{ V_4}^{ V_1} \right) \\ &= \left( \frac{ p_2V_2^{\gamma} }{\gamma - 1} \right) \left( {V_2}^{1- \gamma} - {V_3}^{1- \gamma} \right) + \left( \frac{ p_4V_4^{\gamma} }{\gamma - 1} \right) \left( {V_4}^{1- \gamma} - {V_1}^{1- \gamma} \right) \\ &= \left( \frac{3 p_1V_1}{1 - \gamma} \right) \left( 1 - \frac{1}{4^{\gamma - 1}} \right) - \left( \frac{ p_1V_1}{ \gamma - 1} \right) \left( 1 - \frac{1}{4^{\gamma - 1}} \right) \\ &= \left( \frac{3 nRT_1}{ \gamma - 1} \right) \left( 1 - \frac{1}{4^{\gamma - 1}} \right) - \left( \frac{nRT_1}{ \gamma - 1} \right) \left( 1 - \frac{1}{4^{\gamma - 1}} \right) \\ &= \left( \frac{2 nRT_1}{ \gamma - 1} \right) \left( 1 - \frac{1}{4^{\gamma - 1}} \right) \end{align*}

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