Problem Set 3 Solutions


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1 Chemistry 360 Dr Jean M Standard Problem Set 3 Solutions 1 (a) One mole of an ideal gas at 98 K is expanded reversibly and isothermally from 10 L to 10 L Determine the amount of work in Joules We start with the basic definition of work, Since the process is reversible, P ext P, which gives Substituting the ideal gas equation for P leads to P dv nrt V dv Since the process is isothermal, T can be pulled out of the integral (along with n and R) to yield nrt V 1 V dv nrt ln V / nrt ln( V / ) Now, we are given the initial and final volumes along with the temperature Substituting, ( 1mol) ( 8314 Jmol 1 K 1 ) 98 K 5704 J & ( $ 10 L ' ln% 10 L Notice that the work is negative, indicating work done by the system on the surroundings during the expansion (b) Determine the work done in Joules when one mole of an ideal gas is expanded irreversibly from 10 L to 10 L against a constant external pressure of 10 atm Again, using the definition of work,
2 1 (b) Continued In this case, the external pressure is constant, so it can be pulled out of the integral, P ext V1 V dv Substituting, P ext ( V ) ( 10 L 10 L) P ext V 1 atm 90 L atm Since the result is requested in units of Joules, we must convert units of L atm to Joules w ( 90 L atm) 1013J & $ 1L atm' 91 J Note that work done in this irreversible process is lower in magnitude (even though the initial and final volumes are the same) compared to the work done in part (a) for a reversible process The reversible process always yields the maximum amount of work One mole of an ideal gas at 5 C is expanded reversibly at constant pressure from 0 L to 50 L Determine the amount of work done in Joules We start with the basic definition of work, For a reversible process, the external pressure equals the pressure of the gas, the work becomes P ext P, so the expression for P dv Since the pressure is constant, we can pull it out of the integral and integrate from to V, P V dv P( V )
3 Continued 3 In order to evaluate the work done, the pressure must be calculated We can use the ideal gas equation, along with the initial values of the temperature and volume to determine the pressure, The work done is then given by P nrt K ( 1 mol) L bar mol 1 K 1 P 1394 bar ( 0 L) ( 50 L 0 L) P V 1394 bar 3718 L bar The units of Lbar may then be converted to Joules, 100 J & w ( 3718 L bar) $ 1L bar ' 3718 J 3 (a) One mole of nitrogen gas at 5 C and 1 bar is expanded reversibly and isothermally to a pressure of 013 bar Determine the amount of work done in Joules, assuming ideal behavior We start with the basic definition of work, We saw in problem 1(a) for a reversible isothermal expansion that the work is nrt ln( V / ) Now, we are given the initial and final pressures rather than the initial and final volumes, so the equation above is not directly applicable However, since T is constant, for the initial state we have (from the ideal gas equation) After the expansion, we have for the final state T P 1 nr T P V nr Since the temperatures are equal, we can equate the two relations to give P 1 P V,
4 3 (a) Continued 4 or V P 1 P Substituting this relation into the expression for the work in a reversible isothermal expansion, nrt ln( V / ) Substituting the numerical values yields, w nrt ln P 1 /P or nrt ln( P 1 /P ) ( 1mol) ( 8314 J mol 1 K 1 ) 9815K 500 J ln 10 bar & $ 013 bar ' The work is negative, indicating work done by the system on the surroundings during the expansion (b) Determine the work done when one mole of nitrogen gas at 5 C and 1 bar is expanded isothermally against a constant external pressure of 013 bar assuming ideal behavior This time, the expansion is not reversible, but it is still isothermal The equation for the work is Since the external pressure is constant, it can be pulled outside the integral as in problem 1b and we get P ext V dv P ext ( V ) The external pressure is given in the problem (013 bar), but the initial and final volumes are not given The initial volume can be calculated from the ideal gas equation, nrt P K ( 1mol) L bar mol 1 K L ( 1bar) The final volume can be calculated from the ideal gas equation assuming that the expansion begins with the gas at 1 bar pressure and continues until the pressure of the gas equals the opposing (or external) pressure (since this is the equilibrium state)
5 3 (b) Continued 5 Thus, we assume that P P ext and calculate, V nrt P 9815K ( 1mol) L bar mol 1 K 1 V 1878 L ( 013 bar) Finally, the work can be determined as ( 1878 L 48 L) P ext V 013 bar 15L bar The units of L bar are converted to Joules, 100 J & w ( 15 L bar) $ 1L bar ' 150 J 4 A quantity of 450 g of CaC is reacted at 300 K with excess water in a closed container fitted with a piston, CaC (s) + H O (l) Ca(OH) (aq) + C H (g) Calculate the work done by the production of C H gas against an external pressure of 1 bar Assume that the reaction proceeds until the pressure of the gas produced equals the opposing (or external) pressure The way to start this calculation is to realize that the reaction starts initially with no gas produced; therefore, 0 To get the final volume, assume that the reaction proceeds until the pressure of the gas equals the external pressure Thus, we assume that P P ext and use the ideal gas equation, V nrt P In order to determine the final volume using the equation above, we need to know how many moles of C H are produced Since we started with 450 g CaC, we have " mol %" 450 gcac $ ' 1molC % $ H ' 0070 molc H 6410 gcac & 1molCaC &
6 4 Continued 6 Then, the final volume can be calculated using V nrt P 300 K ( 0070 mol) L bar mol 1 K 1 V 175L ( 1bar) The work for an expansion against constant external pressure can be calculated as ( 175L 0 L) P ext V 1bar 100 J & 175L bar% ( $ 1L bar ' 175J 5 Two moles of O gas, initially at 300 K, are expanded at constant volume from 1 atm to 00 atm Determine the volume of the system and the work done in Joules Under the conditions given, the O gas would be expected to behave ideally (ie, moderate temperature and low pressures) The ideal gas equation may thus be employed to determine the volume of the system The work done is defined as V nrt 1 P K ( mol) L atmmol 1 K 1 V 494 L ( 10 atm) P ext ΔV Since the process takes place at constant volume, and therefore ΔV 0, there is no work done; w 0 6 (a) Derive an expression for the work done in a reversible isothermal expansion of a van der Waals gas from to V We can start from the definition of work, Since the process is reversible, P ext P, and so we have
7 6 (a) Continued P dv 7 An expression for P is required this comes from the van der Waals equation Solving the van der Waals equation for P leads to P nrt V nb an V Substituting this relation into the equation for work yields V nrt V nb an & % $ V ( dv ' V nrt V nb dv + V an dv V Since the process is isothermal, T can be pulled out of the integral along with the other constants to give nrt V nb dv + an V dv Integrating leads to the result nrt ln V nb& an 1 1 & $ nb ' $ V ' (b) One mole of methane expands reversibly from 1 to 50 L at 5 C Calculate the work done assuming (1) the gas behaves ideally, and () the gas obeys the van der Waals equation Part 1: Ideal gas behavior In this case, we can use the result from problem 1 for a reversible isothermal expansion of an ideal gas, nrt ln( V / ) ( 1mol) ( 8314 J mol 1 K 1 ) 9815K 9700 J ln% 50 L $ 1L & ( '
8 6 (b) Continued 8 Part : van der Waals gas behavior For a reversible isothermal expansion of a van der Waals gas, we can use the result developed in part (a) of this question, along with the van der Waals constants for methane (from the textbook) to calculate the work " nrt ln V nb % " $ ' an 1 1 % $ ' nb & V & 9815K ( 0048L mol 1, ) ( 0048L mol 1 ) ( 1mol) 00805L atm mol 1 K 1 ) 50L 1mol ln+ + * 1L 1mol ( 303L atm mol ) 1mol 9675L atm + 6Latm " ( 9449 L atm) 10135J % $ ' 1L atm & 9574J $ 1  " 50L 1 % ' 1L & Note that the work for the expansion of the van der Waals gas is slightly smaller in magnitude than that for the ideal gas The van der Waals gas does less work because some energy is required to overcome the intermolecular interactions in a real gas 7 Derive an expression for the reversible isothermal work from equation of state given by to V done by a gas that obeys the where b is a constant RT, P V m b This problem is very similar to Problem 6(a) Starting from the definition of work, we have Since the process is reversible, P ext P, which leads to P dv In order to evaluate the integral, we need to express P in terms of V Solving the equation of state for P leads to or in terms of V rather than V m, the pressure is P RT V m b,
9 7 Continued P nrt V nb 9 Substituting this relation into the equation for work yields V nrt V nb dv Since the process is isothermal, T can be pulled out of the integral along with the other constants to give nrt V nb dv Integrating leads to the result nrt ln V nb& $ nb ' 8 Determine whether or not the following differentials are exact (a) df xy dx x y dy An exact differential has the form df A( x, y) dx + B( x, y) dy, where % f A x, y & ( $ x ' y and B x, y & ( $ y ' % f x Since the mixed second partial derivatives are equal, the test for exactness is For the differential given above, we have A & $ y ' x B & $ x ' y A x, y Taking the partial derivatives leads to xy and B( x, y) x y A & $ y ' x xy and B & $ x ' y xy Since the partial derivatives are not equal, this differential is NOT EXACT
10 8 Continued 10 (b) df 1 y dx x y dy For this differential, we have Taking the partial derivatives leads to A( x, y) 1 y and B( x, y) x y A & $ y ' x 1 y and B & $ x ' y 1 y Since the partial derivatives are equal, in this case the differential is EXACT 9 Determine whether or not the following differentials are exact (a) dq C v dt + RT V dv (assume that C v and R are constant) In this problem, the differential has the form dq A( T,V ) dt + B( T,V) dv The criterion for exactness is A & $ V ' T B & $ T ' V For the differential given above, we have A T,V Taking the partial derivatives leads to C v and B( T,V) RT V A & $ V ' T 0 and B & $ T ' V R V Since the partial derivatives are not equal, dq is NOT EXACT
11 9 Continued 11 (b) ds C v T dt + R V dv (S is another thermodynamic function called entropy) In this problem, the differential has the form ds A( T,V ) dt + B( T,V) dv The criterion for exactness is A & $ V ' T B & $ T ' V For the differential given above, we have A T,V C v T and B( T,V) R V Taking the partial derivatives leads to A & $ V ' T 0 and B & $ T ' V 0 Since the partial derivatives are equal, the differential ds is EXACT 10 Many compressed gases come in large metal cylinders An 800 L tank of nitrogen gas pressurized to 17 atm is left in the sun and heats from its normal temperature of 00 C to 1400 C Determine (a) the final pressure inside the tank and (b) the work, heat, and ΔU of the process Assume that the nitrogen gas behaves ideally and that the constant volume heat capacity of diatomic nitrogen is 10 J/molK Part (a) To get the final pressure, the ideal gas equation may be employed, P nrt V Everything is known on the right side of this equation except for the number of moles The number of moles can be determined from the initial conditions and the ideal gas equation, n P 1 RT ( 17 atm) 800 L 00805L atm/mol K ( 9315K) n 571 mol
12 10 (a) Continued 1 Now the final pressure may be determined, P nrt V 571mol ( 00805L atm/mol K) ( 41315K) ( 800 L) P 4 atm Part (b) The work is defined as Since there is no change in volume in the process, dv 0 and therefore w0 Assuming that the gas behaves ideally, the differential of internal energy du is defined as du C v dt If C v is constant, we can integrate both sides, ΔU C v dt, or ΔU C v ΔT The molar constant volume heat capacity of nitrogen, C v,m, is given in the problem as 10 J/molK To convert this to heat capacity, C v, it must be multiplied by the number of moles, C v n C v,m 571mol ( 10 J/molK) C v 1014 J/K Substituting, ΔU C v ΔT ( 41315K 9315K) 1014 J /K ΔU 1,441,690 J or 144 kj Now, since w and ΔU are known, the heat q may be calculated from the first law, ΔU q + w We know that w0, so this yields ΔU q Therefore, the heat q is the same as the internal energy change,
13 q ΔU 144 kj 13
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