Two gases NH 3 and phosphine gas, PH 3 are effusing through a small pinhole at same temperature. 1.73 ×10-3 mol NH 3 gas takes time of 11.2 second for effusing. Time taken by PH 3 with same amount that is 1.73 ×10-3 mol is to be determined. Concept introduction: Rate of effusion of is a process of escaping of molecules from high pressure to lower pressure at particular rate. Itcan be calculated by dividing moles of gas by time of effusion. Given as- r a t e = ( n t ) Where n = moles of gas t = effusion time Rate of effusion is also inversely proportional to the square root of the molar mass of atom. Ratio of rate of effusion is given as- R a t e o f e f f u s i o n o f g a s A R a t e o f e f f u s i o n o f g a s B = ( M o l a r m a s s o f g a s B M o l a r m a s s o f g a s A ) 1 / 2

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Question

Chapter 5, Problem 60QAP

Interpretation Introduction

Interpretation:

Two gases NH₃ and phosphine gas, PH₃ are effusing through a small pinhole at same temperature. 1.73 ×10-3 mol NH₃ gas takes time of 11.2 second for effusing. Time taken by PH₃ with same amount that is 1.73 ×10-3 mol is to be determined.

Concept introduction:

Rate of effusion of is a process of escaping of molecules from high pressure to lower pressure at particular rate. Itcan be calculated by dividing moles of gas by time of effusion. Given as-

rate=(nt)

Where n = moles of gas

t = effusion time

Rate of effusion is also inversely proportional to the square root of the molar mass of atom. Ratio of rate of effusion is given as-

Rate of effusion of gas ARate of effusion of gas B=(Molar mass of gas BMolar mass of gas A)1/2

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Answer 2

Question

Chapter 5, Problem 60QAP

Interpretation Introduction

Interpretation:

Two gases NH₃ and phosphine gas, PH₃ are effusing through a small pinhole at same temperature. 1.73 ×10-3 mol NH₃ gas takes time of 11.2 second for effusing. Time taken by PH₃ with same amount that is 1.73 ×10-3 mol is to be determined.

Concept introduction:

Rate of effusion of is a process of escaping of molecules from high pressure to lower pressure at particular rate. Itcan be calculated by dividing moles of gas by time of effusion. Given as-

rate=(nt)

Where n = moles of gas

t = effusion time

Rate of effusion is also inversely proportional to the square root of the molar mass of atom. Ratio of rate of effusion is given as-

Rate of effusion of gas ARate of effusion of gas B=(Molar mass of gas BMolar mass of gas A)1/2

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Answer 3

Question

Chapter 5, Problem 60QAP

Interpretation Introduction

Interpretation:

Two gases NH₃ and phosphine gas, PH₃ are effusing through a small pinhole at same temperature. 1.73 ×10-3 mol NH₃ gas takes time of 11.2 second for effusing. Time taken by PH₃ with same amount that is 1.73 ×10-3 mol is to be determined.

Concept introduction:

Rate of effusion of is a process of escaping of molecules from high pressure to lower pressure at particular rate. Itcan be calculated by dividing moles of gas by time of effusion. Given as-

rate=(nt)

Where n = moles of gas

t = effusion time

Rate of effusion is also inversely proportional to the square root of the molar mass of atom. Ratio of rate of effusion is given as-

Rate of effusion of gas ARate of effusion of gas B=(Molar mass of gas BMolar mass of gas A)1/2

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Answer 4

Question

Chapter 5, Problem 60QAP

Interpretation Introduction

Interpretation:

Two gases NH₃ and phosphine gas, PH₃ are effusing through a small pinhole at same temperature. 1.73 ×10-3 mol NH₃ gas takes time of 11.2 second for effusing. Time taken by PH₃ with same amount that is 1.73 ×10-3 mol is to be determined.

Concept introduction:

Rate of effusion of is a process of escaping of molecules from high pressure to lower pressure at particular rate. Itcan be calculated by dividing moles of gas by time of effusion. Given as-

rate=(nt)

Where n = moles of gas

t = effusion time

Rate of effusion is also inversely proportional to the square root of the molar mass of atom. Ratio of rate of effusion is given as-

Rate of effusion of gas ARate of effusion of gas B=(Molar mass of gas BMolar mass of gas A)1/2

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Answer 5

Chapter 5, Problem 60QAP

Interpretation Introduction

Interpretation:

Two gases NH₃ and phosphine gas, PH₃ are effusing through a small pinhole at same temperature. 1.73 ×10-3 mol NH₃ gas takes time of 11.2 second for effusing. Time taken by PH₃ with same amount that is 1.73 ×10-3 mol is to be determined.

Concept introduction:

Rate of effusion of is a process of escaping of molecules from high pressure to lower pressure at particular rate. Itcan be calculated by dividing moles of gas by time of effusion. Given as-

rate=(nt)

Where n = moles of gas

t = effusion time

Rate of effusion is also inversely proportional to the square root of the molar mass of atom. Ratio of rate of effusion is given as-

Rate of effusion of gas ARate of effusion of gas B=(Molar mass of gas BMolar mass of gas A)1/2

Answer 6

Chapter 5, Problem 60QAP

Interpretation Introduction

Interpretation:

Two gases NH₃ and phosphine gas, PH₃ are effusing through a small pinhole at same temperature. 1.73 ×10-3 mol NH₃ gas takes time of 11.2 second for effusing. Time taken by PH₃ with same amount that is 1.73 ×10-3 mol is to be determined.

Concept introduction:

Rate of effusion of is a process of escaping of molecules from high pressure to lower pressure at particular rate. Itcan be calculated by dividing moles of gas by time of effusion. Given as-

rate=(nt)

Where n = moles of gas

t = effusion time

Rate of effusion is also inversely proportional to the square root of the molar mass of atom. Ratio of rate of effusion is given as-

Rate of effusion of gas ARate of effusion of gas B=(Molar mass of gas BMolar mass of gas A)1/2

Answer 7

Chapter 5, Problem 60QAP

Interpretation Introduction

Interpretation:

Two gases NH₃ and phosphine gas, PH₃ are effusing through a small pinhole at same temperature. 1.73 ×10-3 mol NH₃ gas takes time of 11.2 second for effusing. Time taken by PH₃ with same amount that is 1.73 ×10-3 mol is to be determined.

Concept introduction:

Rate of effusion of is a process of escaping of molecules from high pressure to lower pressure at particular rate. Itcan be calculated by dividing moles of gas by time of effusion. Given as-

rate=(nt)

Where n = moles of gas

t = effusion time

Rate of effusion is also inversely proportional to the square root of the molar mass of atom. Ratio of rate of effusion is given as-

Rate of effusion of gas ARate of effusion of gas B=(Molar mass of gas BMolar mass of gas A)1/2

Answer 8

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Answer 9

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Answer 11

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Solution Summary: The author explains that rate of effusion is a process of escaping molecules from high pressure to lower pressure at particular rate.

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