The number of moles per second of purified hydrogen that can be produced.

Question

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Answer 1

Question

Chapter 4, Problem 4.21P

To determine

The number of moles per second of purified hydrogen that can be produced.

Expert Solution & Answer

Answer to Problem 4.21P

The number of moles per second of purified hydrogen that can be produced is 2.243×10−2 moles/s.

Explanation of Solution

Given:

The hydrogen purifier is made of a tube of palladium with 25 weight percent silver.

The inner radius of the tube is 5.11×10−4 m.

The outer radius is 6.35×10−4 m.

The length of the tube is 5.81 m.

The inside high pressure is 1×106 N/m2.

The outside low pressure is 1×104 N/m2.

The temperature is 673 K.

Formula used:

The flux of hydrogen ions through the tube of palladium is given by,

JH=[4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] ....... (I)

Here, JH is the flux of hydrogen ions, ph insider high pressure, ph is the outside high pressure, r2 is the outer radius, r1 is the inner radius, R is the universal gas constant and T is the temperature.

The effective area of the tube is given by,

A=2πl(r2−r1)ln( r 2 r 1 ) ....... (II)

Here, A is the effective area of the tube and l is the length of the tube.

The hydrogen ions produced per second is given by,

(JH)PS=JHA ....... (III)

Here, (JH)PS are the hydrogen ions produced per second.

Substitute [4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] for JH and 2πl(r2−r1)ln( r 2 r 1 ) for A in equation (III).

(JH)PS=[[4.86× 10 −7( p h 1 2 − p l 1 2 r 2 − r 1 ) moles m 2 ⋅s][exp( −( 6.6× 10 3 J/ mole RT ))][ 2πl( r 2 − r 1 ) ln( r 2 r 1 )]] ....... (IV)

Calculation:

The hydrogen ions produced per second is calculated as,

Substitute 1×106 N/m2 for ph, 1×104 N/m2 for pl, 6.35×10−4 m for r2, 5.11×10−4 m for r1, 673 K for T, 5.81 m for l and 8.31 J/mole⋅K for R in equation (IV).

( J H ) PS =[ [ 4.86× 10 −7 { ( 1× 10 6 N/ m 2 ) 1 2 − ( 1× 10 4 N/ m 2 ) 1 2 ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } moles m 2 ⋅s ] [ exp{ −( 6.6× 10 3 J/ mole ( 8.31 J/ mole⋅K )( 673 K ) ) } ] [ 2π( 5.81 m ){ ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } ln( 6.35× 10 −4 m 5.11× 10 −4 m ) ] ]

=[ [ 4.86× 10 −7 ( 900 1.24× 10 −4 ) moles m 2 ⋅s ] [ exp( −( 6.6× 10 3 5592.63 ) ) ][ 45.27× 10 −4 m 2 0.219 ] ]

=[ [ 3.527 moles m 2 ⋅s ][ 0.3072 ][ 0.0207 m 2 ] ]

=2.243× 10 −2 moles/s

Conclusion:

Therefore, the number of moles per second of purified hydrogen that can be produced is 2.243×10−2 moles/s.

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Answer 2

Question

Chapter 4, Problem 4.21P

To determine

The number of moles per second of purified hydrogen that can be produced.

Expert Solution & Answer

Answer to Problem 4.21P

The number of moles per second of purified hydrogen that can be produced is 2.243×10−2 moles/s.

Explanation of Solution

Given:

The hydrogen purifier is made of a tube of palladium with 25 weight percent silver.

The inner radius of the tube is 5.11×10−4 m.

The outer radius is 6.35×10−4 m.

The length of the tube is 5.81 m.

The inside high pressure is 1×106 N/m2.

The outside low pressure is 1×104 N/m2.

The temperature is 673 K.

Formula used:

The flux of hydrogen ions through the tube of palladium is given by,

JH=[4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] ....... (I)

Here, JH is the flux of hydrogen ions, ph insider high pressure, ph is the outside high pressure, r2 is the outer radius, r1 is the inner radius, R is the universal gas constant and T is the temperature.

The effective area of the tube is given by,

A=2πl(r2−r1)ln( r 2 r 1 ) ....... (II)

Here, A is the effective area of the tube and l is the length of the tube.

The hydrogen ions produced per second is given by,

(JH)PS=JHA ....... (III)

Here, (JH)PS are the hydrogen ions produced per second.

Substitute [4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] for JH and 2πl(r2−r1)ln( r 2 r 1 ) for A in equation (III).

(JH)PS=[[4.86× 10 −7( p h 1 2 − p l 1 2 r 2 − r 1 ) moles m 2 ⋅s][exp( −( 6.6× 10 3 J/ mole RT ))][ 2πl( r 2 − r 1 ) ln( r 2 r 1 )]] ....... (IV)

Calculation:

The hydrogen ions produced per second is calculated as,

Substitute 1×106 N/m2 for ph, 1×104 N/m2 for pl, 6.35×10−4 m for r2, 5.11×10−4 m for r1, 673 K for T, 5.81 m for l and 8.31 J/mole⋅K for R in equation (IV).

( J H ) PS =[ [ 4.86× 10 −7 { ( 1× 10 6 N/ m 2 ) 1 2 − ( 1× 10 4 N/ m 2 ) 1 2 ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } moles m 2 ⋅s ] [ exp{ −( 6.6× 10 3 J/ mole ( 8.31 J/ mole⋅K )( 673 K ) ) } ] [ 2π( 5.81 m ){ ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } ln( 6.35× 10 −4 m 5.11× 10 −4 m ) ] ]

=[ [ 4.86× 10 −7 ( 900 1.24× 10 −4 ) moles m 2 ⋅s ] [ exp( −( 6.6× 10 3 5592.63 ) ) ][ 45.27× 10 −4 m 2 0.219 ] ]

=[ [ 3.527 moles m 2 ⋅s ][ 0.3072 ][ 0.0207 m 2 ] ]

=2.243× 10 −2 moles/s

Conclusion:

Therefore, the number of moles per second of purified hydrogen that can be produced is 2.243×10−2 moles/s.

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Answer 3

Question

Chapter 4, Problem 4.21P

To determine

The number of moles per second of purified hydrogen that can be produced.

Expert Solution & Answer

Answer to Problem 4.21P

The number of moles per second of purified hydrogen that can be produced is 2.243×10−2 moles/s.

Explanation of Solution

Given:

The hydrogen purifier is made of a tube of palladium with 25 weight percent silver.

The inner radius of the tube is 5.11×10−4 m.

The outer radius is 6.35×10−4 m.

The length of the tube is 5.81 m.

The inside high pressure is 1×106 N/m2.

The outside low pressure is 1×104 N/m2.

The temperature is 673 K.

Formula used:

The flux of hydrogen ions through the tube of palladium is given by,

JH=[4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] ....... (I)

Here, JH is the flux of hydrogen ions, ph insider high pressure, ph is the outside high pressure, r2 is the outer radius, r1 is the inner radius, R is the universal gas constant and T is the temperature.

The effective area of the tube is given by,

A=2πl(r2−r1)ln( r 2 r 1 ) ....... (II)

Here, A is the effective area of the tube and l is the length of the tube.

The hydrogen ions produced per second is given by,

(JH)PS=JHA ....... (III)

Here, (JH)PS are the hydrogen ions produced per second.

Substitute [4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] for JH and 2πl(r2−r1)ln( r 2 r 1 ) for A in equation (III).

(JH)PS=[[4.86× 10 −7( p h 1 2 − p l 1 2 r 2 − r 1 ) moles m 2 ⋅s][exp( −( 6.6× 10 3 J/ mole RT ))][ 2πl( r 2 − r 1 ) ln( r 2 r 1 )]] ....... (IV)

Calculation:

The hydrogen ions produced per second is calculated as,

Substitute 1×106 N/m2 for ph, 1×104 N/m2 for pl, 6.35×10−4 m for r2, 5.11×10−4 m for r1, 673 K for T, 5.81 m for l and 8.31 J/mole⋅K for R in equation (IV).

( J H ) PS =[ [ 4.86× 10 −7 { ( 1× 10 6 N/ m 2 ) 1 2 − ( 1× 10 4 N/ m 2 ) 1 2 ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } moles m 2 ⋅s ] [ exp{ −( 6.6× 10 3 J/ mole ( 8.31 J/ mole⋅K )( 673 K ) ) } ] [ 2π( 5.81 m ){ ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } ln( 6.35× 10 −4 m 5.11× 10 −4 m ) ] ]

=[ [ 4.86× 10 −7 ( 900 1.24× 10 −4 ) moles m 2 ⋅s ] [ exp( −( 6.6× 10 3 5592.63 ) ) ][ 45.27× 10 −4 m 2 0.219 ] ]

=[ [ 3.527 moles m 2 ⋅s ][ 0.3072 ][ 0.0207 m 2 ] ]

=2.243× 10 −2 moles/s

Conclusion:

Therefore, the number of moles per second of purified hydrogen that can be produced is 2.243×10−2 moles/s.

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Answer 4

Question

Chapter 4, Problem 4.21P

To determine

The number of moles per second of purified hydrogen that can be produced.

Expert Solution & Answer

Answer to Problem 4.21P

The number of moles per second of purified hydrogen that can be produced is 2.243×10−2 moles/s.

Explanation of Solution

Given:

The hydrogen purifier is made of a tube of palladium with 25 weight percent silver.

The inner radius of the tube is 5.11×10−4 m.

The outer radius is 6.35×10−4 m.

The length of the tube is 5.81 m.

The inside high pressure is 1×106 N/m2.

The outside low pressure is 1×104 N/m2.

The temperature is 673 K.

Formula used:

The flux of hydrogen ions through the tube of palladium is given by,

JH=[4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] ....... (I)

Here, JH is the flux of hydrogen ions, ph insider high pressure, ph is the outside high pressure, r2 is the outer radius, r1 is the inner radius, R is the universal gas constant and T is the temperature.

The effective area of the tube is given by,

A=2πl(r2−r1)ln( r 2 r 1 ) ....... (II)

Here, A is the effective area of the tube and l is the length of the tube.

The hydrogen ions produced per second is given by,

(JH)PS=JHA ....... (III)

Here, (JH)PS are the hydrogen ions produced per second.

Substitute [4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] for JH and 2πl(r2−r1)ln( r 2 r 1 ) for A in equation (III).

(JH)PS=[[4.86× 10 −7( p h 1 2 − p l 1 2 r 2 − r 1 ) moles m 2 ⋅s][exp( −( 6.6× 10 3 J/ mole RT ))][ 2πl( r 2 − r 1 ) ln( r 2 r 1 )]] ....... (IV)

Calculation:

The hydrogen ions produced per second is calculated as,

Substitute 1×106 N/m2 for ph, 1×104 N/m2 for pl, 6.35×10−4 m for r2, 5.11×10−4 m for r1, 673 K for T, 5.81 m for l and 8.31 J/mole⋅K for R in equation (IV).

( J H ) PS =[ [ 4.86× 10 −7 { ( 1× 10 6 N/ m 2 ) 1 2 − ( 1× 10 4 N/ m 2 ) 1 2 ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } moles m 2 ⋅s ] [ exp{ −( 6.6× 10 3 J/ mole ( 8.31 J/ mole⋅K )( 673 K ) ) } ] [ 2π( 5.81 m ){ ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } ln( 6.35× 10 −4 m 5.11× 10 −4 m ) ] ]

=[ [ 4.86× 10 −7 ( 900 1.24× 10 −4 ) moles m 2 ⋅s ] [ exp( −( 6.6× 10 3 5592.63 ) ) ][ 45.27× 10 −4 m 2 0.219 ] ]

=[ [ 3.527 moles m 2 ⋅s ][ 0.3072 ][ 0.0207 m 2 ] ]

=2.243× 10 −2 moles/s

Conclusion:

Therefore, the number of moles per second of purified hydrogen that can be produced is 2.243×10−2 moles/s.

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Answer 5

Chapter 4, Problem 4.21P

To determine

The number of moles per second of purified hydrogen that can be produced.

Expert Solution & Answer

Answer to Problem 4.21P

The number of moles per second of purified hydrogen that can be produced is 2.243×10−2 moles/s.

Explanation of Solution

Given:

The hydrogen purifier is made of a tube of palladium with 25 weight percent silver.

The inner radius of the tube is 5.11×10−4 m.

The outer radius is 6.35×10−4 m.

The length of the tube is 5.81 m.

The inside high pressure is 1×106 N/m2.

The outside low pressure is 1×104 N/m2.

The temperature is 673 K.

Formula used:

The flux of hydrogen ions through the tube of palladium is given by,

JH=[4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] ....... (I)

Here, JH is the flux of hydrogen ions, ph insider high pressure, ph is the outside high pressure, r2 is the outer radius, r1 is the inner radius, R is the universal gas constant and T is the temperature.

The effective area of the tube is given by,

A=2πl(r2−r1)ln( r 2 r 1 ) ....... (II)

Here, A is the effective area of the tube and l is the length of the tube.

The hydrogen ions produced per second is given by,

(JH)PS=JHA ....... (III)

Here, (JH)PS are the hydrogen ions produced per second.

Substitute [4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] for JH and 2πl(r2−r1)ln( r 2 r 1 ) for A in equation (III).

(JH)PS=[[4.86× 10 −7( p h 1 2 − p l 1 2 r 2 − r 1 ) moles m 2 ⋅s][exp( −( 6.6× 10 3 J/ mole RT ))][ 2πl( r 2 − r 1 ) ln( r 2 r 1 )]] ....... (IV)

Calculation:

The hydrogen ions produced per second is calculated as,

Substitute 1×106 N/m2 for ph, 1×104 N/m2 for pl, 6.35×10−4 m for r2, 5.11×10−4 m for r1, 673 K for T, 5.81 m for l and 8.31 J/mole⋅K for R in equation (IV).

( J H ) PS =[ [ 4.86× 10 −7 { ( 1× 10 6 N/ m 2 ) 1 2 − ( 1× 10 4 N/ m 2 ) 1 2 ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } moles m 2 ⋅s ] [ exp{ −( 6.6× 10 3 J/ mole ( 8.31 J/ mole⋅K )( 673 K ) ) } ] [ 2π( 5.81 m ){ ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } ln( 6.35× 10 −4 m 5.11× 10 −4 m ) ] ]

=[ [ 4.86× 10 −7 ( 900 1.24× 10 −4 ) moles m 2 ⋅s ] [ exp( −( 6.6× 10 3 5592.63 ) ) ][ 45.27× 10 −4 m 2 0.219 ] ]

=[ [ 3.527 moles m 2 ⋅s ][ 0.3072 ][ 0.0207 m 2 ] ]

=2.243× 10 −2 moles/s

Conclusion:

Therefore, the number of moles per second of purified hydrogen that can be produced is 2.243×10−2 moles/s.

Answer 6

Chapter 4, Problem 4.21P

To determine

The number of moles per second of purified hydrogen that can be produced.

Expert Solution & Answer

Answer to Problem 4.21P

The number of moles per second of purified hydrogen that can be produced is 2.243×10−2 moles/s.

Explanation of Solution

Given:

The hydrogen purifier is made of a tube of palladium with 25 weight percent silver.

The inner radius of the tube is 5.11×10−4 m.

The outer radius is 6.35×10−4 m.

The length of the tube is 5.81 m.

The inside high pressure is 1×106 N/m2.

The outside low pressure is 1×104 N/m2.

The temperature is 673 K.

Formula used:

The flux of hydrogen ions through the tube of palladium is given by,

JH=[4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] ....... (I)

Here, JH is the flux of hydrogen ions, ph insider high pressure, ph is the outside high pressure, r2 is the outer radius, r1 is the inner radius, R is the universal gas constant and T is the temperature.

The effective area of the tube is given by,

A=2πl(r2−r1)ln( r 2 r 1 ) ....... (II)

Here, A is the effective area of the tube and l is the length of the tube.

The hydrogen ions produced per second is given by,

(JH)PS=JHA ....... (III)

Here, (JH)PS are the hydrogen ions produced per second.

Substitute [4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] for JH and 2πl(r2−r1)ln( r 2 r 1 ) for A in equation (III).

(JH)PS=[[4.86× 10 −7( p h 1 2 − p l 1 2 r 2 − r 1 ) moles m 2 ⋅s][exp( −( 6.6× 10 3 J/ mole RT ))][ 2πl( r 2 − r 1 ) ln( r 2 r 1 )]] ....... (IV)

Calculation:

The hydrogen ions produced per second is calculated as,

Substitute 1×106 N/m2 for ph, 1×104 N/m2 for pl, 6.35×10−4 m for r2, 5.11×10−4 m for r1, 673 K for T, 5.81 m for l and 8.31 J/mole⋅K for R in equation (IV).

( J H ) PS =[ [ 4.86× 10 −7 { ( 1× 10 6 N/ m 2 ) 1 2 − ( 1× 10 4 N/ m 2 ) 1 2 ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } moles m 2 ⋅s ] [ exp{ −( 6.6× 10 3 J/ mole ( 8.31 J/ mole⋅K )( 673 K ) ) } ] [ 2π( 5.81 m ){ ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } ln( 6.35× 10 −4 m 5.11× 10 −4 m ) ] ]

=[ [ 4.86× 10 −7 ( 900 1.24× 10 −4 ) moles m 2 ⋅s ] [ exp( −( 6.6× 10 3 5592.63 ) ) ][ 45.27× 10 −4 m 2 0.219 ] ]

=[ [ 3.527 moles m 2 ⋅s ][ 0.3072 ][ 0.0207 m 2 ] ]

=2.243× 10 −2 moles/s

Conclusion:

Therefore, the number of moles per second of purified hydrogen that can be produced is 2.243×10−2 moles/s.

Answer 7

Chapter 4, Problem 4.21P

To determine

The number of moles per second of purified hydrogen that can be produced.

Answer 8

Expert Solution & Answer

Answer to Problem 4.21P

The number of moles per second of purified hydrogen that can be produced is 2.243×10−2 moles/s.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

The hydrogen purifier is made of a tube of palladium with 25 weight percent silver.

The inner radius of the tube is 5.11×10−4 m.

The outer radius is 6.35×10−4 m.

The length of the tube is 5.81 m.

The inside high pressure is 1×106 N/m2.

The outside low pressure is 1×104 N/m2.

The temperature is 673 K.

Formula used:

The flux of hydrogen ions through the tube of palladium is given by,

JH=[4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] ....... (I)

Here, JH is the flux of hydrogen ions, ph insider high pressure, ph is the outside high pressure, r2 is the outer radius, r1 is the inner radius, R is the universal gas constant and T is the temperature.

The effective area of the tube is given by,

A=2πl(r2−r1)ln( r 2 r 1 ) ....... (II)

Here, A is the effective area of the tube and l is the length of the tube.

The hydrogen ions produced per second is given by,

(JH)PS=JHA ....... (III)

Here, (JH)PS are the hydrogen ions produced per second.

Substitute [4.86×10−7( p h 1 2 − p l 1 2 r 2− r 1)molesm2⋅s][exp(−( 6.6× 10 3 J/ mole RT))] for JH and 2πl(r2−r1)ln( r 2 r 1 ) for A in equation (III).

(JH)PS=[[4.86× 10 −7( p h 1 2 − p l 1 2 r 2 − r 1 ) moles m 2 ⋅s][exp( −( 6.6× 10 3 J/ mole RT ))][ 2πl( r 2 − r 1 ) ln( r 2 r 1 )]] ....... (IV)

Calculation:

The hydrogen ions produced per second is calculated as,

Substitute 1×106 N/m2 for ph, 1×104 N/m2 for pl, 6.35×10−4 m for r2, 5.11×10−4 m for r1, 673 K for T, 5.81 m for l and 8.31 J/mole⋅K for R in equation (IV).

( J H ) PS =[ [ 4.86× 10 −7 { ( 1× 10 6 N/ m 2 ) 1 2 − ( 1× 10 4 N/ m 2 ) 1 2 ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } moles m 2 ⋅s ] [ exp{ −( 6.6× 10 3 J/ mole ( 8.31 J/ mole⋅K )( 673 K ) ) } ] [ 2π( 5.81 m ){ ( 6.35× 10 −4 m )−( 5.11× 10 −4 m ) } ln( 6.35× 10 −4 m 5.11× 10 −4 m ) ] ]