PS4-Solution_08 - Chem 444B Homework Set #4 - Solutions...

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Problem Set #4 -1- Chem 444B Homework Set #4 - Solutions DUE: Friday, February 15, 2008 1) A chemical reaction takes place in a container of cross-sectional area 100 cm 2 . As a result of the reaction, a piston is pushed out through 10 cm against an external pressure of 1.0 atm. Calculate the work (in J) done. Was this work done on the system or by the system ? () 23 Since the external pressure is constant, work is given by Eq. 19.1: 100 10 1000 1000 1 1 ext wP V Vc m c m c m m L L wa t m =− ∆= = = = 5 1.01 10 1 Pa atm × × 1 L ⎛⎞ ⎜⎟ ⎝⎠ 3 1 1000 m L × 3 101 101 Pa m wJ Since the value of work is less than 0, work was done by the system . 2) a) A sample of 4.50 g of methane occupies 12.7 L at 310K. Calculate the work (in J) done when the gas expands isothermally against a constant external pressure of 7.7 kPa until its volume has increased by 3.3 L. 3 Since the external pressure is constant, work is given by Eq. 19.1: 7.7 10 3.3 ext V a L × 3 1 1000 m L × 3 25 25 Pa m b) Calculate the work done if the same expansion had occurred reversibly. 22 2 11 1 2 1 4 44 4 4 Since the pressure is NOT constant, reversible work is given by Eq. 19.4: ln 1 4.50 0.28 16 0.28 8.314 VV V rev gas V rev V nRT dV w P dV dV nRT nRT V mol CH ng C H m o l C H gCH J wm o l C H mol K = ∫∫ (12.7 3.3) 310 ln 12.7 166.7 L K L +
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Problem Set #4 -2- 3) a) Repeat problem 2b for a van der Waals gas using the expression derived in Lecture 10 () 62 3 22 2 2 11 2 4 Use Table 16.3 to find the values of a and b for methane: 2.3026 =2.3026 0.043067 0.043067 ln 0.28 0 rev rev dm bar L bar dm L ab mol mol mol mol Vn b wn R T a n b VV wm o l C H == = ⎛⎞ =− + ⎜⎟ ⎝⎠ ( ) 2 2 2 16.0 0.28 0.043067 .083145 310 ln 12.7 0.28 0.043067 2.3026 0.28 12.7 16.0 7.217 ln 1.260 2. rev L Lm o l L bar mol K L mol K o l mol L bar mol mol L L w L bar +− + 3 53 93 10 10 1 1.666 1 1000 166.6 rev rev L bar Pa m w L bar bar L wJ ×⋅ × × b) Derive an expression for the work done during a reversible isothermal expansion of a Redlich-Kwong gas 1/2 2 2 1/ 1 Begin with Eq. 19.4 but use the expression for a Redlich-Kwong gas (Eq. 16.7): ln V V rev gas V V rev nRT An dV An dV w P dV dV nRT BTVVn B BT VVn B B A n R T B T ⎡⎤ + ⎢⎥ −+ + ⎣⎦ + ∫∫ 21 2 1 2 2 2 2 1 1 ln ln( ) ln ln ln ln rev V n BV n B nB nB nB nB VV n B n R T B TB n VVn B
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This homework help was uploaded on 04/13/2008 for the course CHEM 444 taught by Professor Jameslis during the Spring '08 term at University of Illinois at Urbana–Champaign.

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PS4-Solution_08 - Chem 444B Homework Set #4 - Solutions...

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