Pressure exerted by 1.00 mol C 2 H 6 when it behaves as ideal gas has to be determined. Concept Introduction: The expression for ideal gas equation for one mole of gas is as follows: P V = R T (1) Here, P is pressure of gas. V is molar volume of gas. T is absolute temperature of gas. R is universal gas constant.

Question

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Answer 1

Question

Chapter 3, Problem 3E.10E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as ideal gas has to be determined.

Concept Introduction:

The expression for ideal gas equation for one mole of gas is as follows:

PV=RT (1)

Here,

P is pressure of gas.

V is molar volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (2)

(i) Substitute 1.121 L for V, 273.15 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)1.121 L=20.26 bar

(ii) Substitute 40.0 L for V, 1.00×103 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)40.0 L=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by ideal gas is 20.26 bar and 2.08 bar.

(b)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as van der Waals gas has to be determined.

Concept Introduction:

The formula to calculate pressure of one mole of gas is as follows:

P=RT(V−b)−aV2 (3)

Here,

P is pressure of gas.

V is volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

a and b are van der Waals parameters.

(b)

Expert Solution

Explanation of Solution

(i) Substitute 1.121 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)(1.121 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(1.121 L)2)=17.12 bar

(ii) Substitute 40.0 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)(40.0 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(40.0 L)2)=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by van der Waals gas is 17.12 bar and 2.08 bar.

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Answer 2

Question

Chapter 3, Problem 3E.10E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as ideal gas has to be determined.

Concept Introduction:

The expression for ideal gas equation for one mole of gas is as follows:

PV=RT (1)

Here,

P is pressure of gas.

V is molar volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (2)

(i) Substitute 1.121 L for V, 273.15 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)1.121 L=20.26 bar

(ii) Substitute 40.0 L for V, 1.00×103 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)40.0 L=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by ideal gas is 20.26 bar and 2.08 bar.

(b)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as van der Waals gas has to be determined.

Concept Introduction:

The formula to calculate pressure of one mole of gas is as follows:

P=RT(V−b)−aV2 (3)

Here,

P is pressure of gas.

V is volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

a and b are van der Waals parameters.

(b)

Expert Solution

Explanation of Solution

(i) Substitute 1.121 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)(1.121 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(1.121 L)2)=17.12 bar

(ii) Substitute 40.0 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)(40.0 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(40.0 L)2)=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by van der Waals gas is 17.12 bar and 2.08 bar.

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Answer 3

Question

Chapter 3, Problem 3E.10E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as ideal gas has to be determined.

Concept Introduction:

The expression for ideal gas equation for one mole of gas is as follows:

PV=RT (1)

Here,

P is pressure of gas.

V is molar volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (2)

(i) Substitute 1.121 L for V, 273.15 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)1.121 L=20.26 bar

(ii) Substitute 40.0 L for V, 1.00×103 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)40.0 L=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by ideal gas is 20.26 bar and 2.08 bar.

(b)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as van der Waals gas has to be determined.

Concept Introduction:

The formula to calculate pressure of one mole of gas is as follows:

P=RT(V−b)−aV2 (3)

Here,

P is pressure of gas.

V is volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

a and b are van der Waals parameters.

(b)

Expert Solution

Explanation of Solution

(i) Substitute 1.121 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)(1.121 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(1.121 L)2)=17.12 bar

(ii) Substitute 40.0 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)(40.0 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(40.0 L)2)=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by van der Waals gas is 17.12 bar and 2.08 bar.

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Answer 4

Question

Chapter 3, Problem 3E.10E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as ideal gas has to be determined.

Concept Introduction:

The expression for ideal gas equation for one mole of gas is as follows:

PV=RT (1)

Here,

P is pressure of gas.

V is molar volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (2)

(i) Substitute 1.121 L for V, 273.15 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)1.121 L=20.26 bar

(ii) Substitute 40.0 L for V, 1.00×103 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)40.0 L=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by ideal gas is 20.26 bar and 2.08 bar.

(b)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as van der Waals gas has to be determined.

Concept Introduction:

The formula to calculate pressure of one mole of gas is as follows:

P=RT(V−b)−aV2 (3)

Here,

P is pressure of gas.

V is volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

a and b are van der Waals parameters.

(b)

Expert Solution

Explanation of Solution

(i) Substitute 1.121 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)(1.121 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(1.121 L)2)=17.12 bar

(ii) Substitute 40.0 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)(40.0 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(40.0 L)2)=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by van der Waals gas is 17.12 bar and 2.08 bar.

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Answer 5

Chapter 3, Problem 3E.10E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as ideal gas has to be determined.

Concept Introduction:

The expression for ideal gas equation for one mole of gas is as follows:

PV=RT (1)

Here,

P is pressure of gas.

V is molar volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (2)

(i) Substitute 1.121 L for V, 273.15 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)1.121 L=20.26 bar

(ii) Substitute 40.0 L for V, 1.00×103 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)40.0 L=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by ideal gas is 20.26 bar and 2.08 bar.

(b)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as van der Waals gas has to be determined.

Concept Introduction:

The formula to calculate pressure of one mole of gas is as follows:

P=RT(V−b)−aV2 (3)

Here,

P is pressure of gas.

V is volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

a and b are van der Waals parameters.

(b)

Expert Solution

Explanation of Solution

(i) Substitute 1.121 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)(1.121 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(1.121 L)2)=17.12 bar

(ii) Substitute 40.0 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)(40.0 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(40.0 L)2)=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by van der Waals gas is 17.12 bar and 2.08 bar.

Answer 6

Chapter 3, Problem 3E.10E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as ideal gas has to be determined.

Concept Introduction:

The expression for ideal gas equation for one mole of gas is as follows:

PV=RT (1)

Here,

P is pressure of gas.

V is molar volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (2)

(i) Substitute 1.121 L for V, 273.15 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)1.121 L=20.26 bar

(ii) Substitute 40.0 L for V, 1.00×103 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)40.0 L=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by ideal gas is 20.26 bar and 2.08 bar.

Answer 7

Chapter 3, Problem 3E.10E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as ideal gas has to be determined.

Concept Introduction:

The expression for ideal gas equation for one mole of gas is as follows:

PV=RT (1)

Here,

P is pressure of gas.

V is molar volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

Answer 8

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (2)

(i) Substitute 1.121 L for V, 273.15 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)1.121 L=20.26 bar

(ii) Substitute 40.0 L for V, 1.00×103 K for T, and 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R in equation (2).

P=(8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)40.0 L=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by ideal gas is 20.26 bar and 2.08 bar.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as van der Waals gas has to be determined.

Concept Introduction:

The formula to calculate pressure of one mole of gas is as follows:

P=RT(V−b)−aV2 (3)

Here,

P is pressure of gas.

V is volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

a and b are van der Waals parameters.

(b)

Expert Solution

Explanation of Solution

(i) Substitute 1.121 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)(1.121 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(1.121 L)2)=17.12 bar

(ii) Substitute 40.0 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)(40.0 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(40.0 L)2)=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by van der Waals gas is 17.12 bar and 2.08 bar.

Answer 13

(b)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol C2H6 when it behaves as van der Waals gas has to be determined.

Concept Introduction:

The formula to calculate pressure of one mole of gas is as follows:

P=RT(V−b)−aV2 (3)

Here,

P is pressure of gas.

V is volume of gas.

T is absolute temperature of gas.

R is universal gas constant.

a and b are van der Waals parameters.

Answer 14

(b)

Expert Solution

Explanation of Solution

(i) Substitute 1.121 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(273.15 K)(1.121 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(1.121 L)2)=17.12 bar

(ii) Substitute 40.0 L for V 273.15 K for T, 8.3145×10−2 L⋅bar⋅K−1⋅mol−1 for R, 5.507 bar⋅L2⋅mol−2 for a and 6.51×10−2 L⋅mol−1 for b in equation (3).

P=((8.3145×10−2 L⋅bar⋅K−1⋅mol−1)(1.00×103 K)(40.0 L−6.51×10−2 L⋅mol−1))−(5.507 bar⋅L2⋅mol−2(40.0 L)2)=2.08 bar

Pressure exerted by 1.00 mol C2H6 in case (i) and (ii) by van der Waals gas is 17.12 bar and 2.08 bar.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Ch. 3 - Prob. 3A.1AST Ch. 3 - Prob. 3A.1BST Ch. 3 - Prob. 3A.2AST Ch. 3 - Prob. 3A.2BST Ch. 3 - Prob. 3A.3AST Ch. 3 - Prob. 3A.3BST Ch. 3 - Prob. 3A.1E Ch. 3 - Prob. 3A.2E Ch. 3 - Prob. 3A.3E Ch. 3 - Prob. 3A.4E

Explanation of Solution

Explanation of Solution

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Chapter 3 Solutions