Chem 2B-quiz 3-011907

Chem 2B-quiz 3-011907 - proportion; E = nc v T T = E/ nc v...

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Chem 2B, Winter 2007 Professor Thuc-Quyen Nguyen Name: Quiz #3 01/19/07 1) ( 5 points ) Suppose 2.00 mol of an ideal, monoatomic gas is initially at a pressure of 3.00 atm and a temperature T = 350 K. It is expanded irreversibly and adiabatically (q=0) against a constant external pressure of 1.00 atm until the volume has doubled. (a) Calculate the final volume. Initial volume V = nRT/P = 19.15 L The final volume is twice this original volume = 38.3 L (b) Calculate w, q, and E for this process in joules. Adiabatic expansion occurs against a const. P w = -PΔV = -1.00(19.15)L atm x (101.325 J/1 L atm) = -1.94x10 3 J The expansion is adiabatic, q = 0; ΔE = q + w = 0 - 1.94x10 3 J = -1.94x10 3 J (c) Calculate the final temperature of the gas. Any change in the internal energy of an ideal gas causes a change in temperature in direct
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Unformatted text preview: proportion; E = nc v T T = E/ nc v = -1.94x10 3 J/[(2.00 mol)(3/2)(8.3145 J K-1 mol-1 )] = -77.8K Final temperature T 2 = T 1 + T = 350 + (-77.8) = 272 K 2) ( 5 points ) In walking a kilometer, you use about 100 kJ of energy. This energy comes from the oxidation of foods, which is about 30% efficient. How much energy do you save by walking a kilometer instead of driving a car that gets 8.0 km/L of gasoline (19 miles per gallon)? The density of gasoline is 0.68 g/cm 3 and its enthalpy of combustion is -48 kJ/g. The energy released in the combustion of gasoline in fueling the car to travel a kilometer; 1km x (1 L gasoline/8.0 km) x (0.68 kg/1 L) x (48x10 3 kJ/ 1 kg) = 4080 kJ The energy released in he oxidation of food is about (100/0.3)kJ = 330 kJ The energy savings is ~ 3750 kJ...
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This note was uploaded on 08/06/2008 for the course CHEM 2B taught by Professor Nguyen during the Winter '07 term at UCSB.

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