(a) Interpretation: To find the heat liberated of isothermal mixing at 300 K Concept Introduction: We find the Excess enthalpy of H 2 SO 4 of pure and 25-wt-% from the H-x diagram of H 2 SO 4 Then we find the enthalpy of Pure water. The heat of mixing, Q is as given below, Q = x 4 H 4 − x 1 H 1 − x 2 H 2 − x 3 H 3 Where, Mass of H 2 O = m 1 Mass of H 2 SO 4 = m 2 Mass of 25-wt-% H 2 SO 4 = m 3 Total mass = m 4 Enthalpy of Pure H 2 O = H 1 Enthalpy of Pure H 2 SO 4 (100-wt-%) = H 2 Enthalpy of 25-wt-% H 2 SO 4 = H 3 Total enthalpy = H 4 Mass fraction of H 2 O = x 1 Mass fraction of H 2 SO 4 = x 2 Mass fraction of 25-wt-% H 2 SO 4 = x 3 Total mass fraction of H 2 SO 4 = x 4

Question

Want to see the full answer?

Answer 1

Question

Chapter 11, Problem 11.30P

Interpretation Introduction

(a)

Interpretation:

To find the heat liberated of isothermal mixing at 300 K

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x4H4−x1H1−x2H2−x3H3

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Mass of 25-wt-% H₂SO₄= m₃ Total mass = m₄

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 25-wt-% H₂SO₄ = H₃ Total enthalpy = H₄

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Mass fraction of 25-wt-% H₂SO₄= x₃ Total mass fraction of H₂SO₄= x₄

(b)

Interpretation Introduction

Interpretation:

To find the temperature of the intermediate solution formed

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x3H3−x1H1−x2H2

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Total mass = m₃

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Total enthalpy = H₃

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Total mass fraction of H₂SO₄= x₃

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 2

Question

Chapter 11, Problem 11.30P

Interpretation Introduction

(a)

Interpretation:

To find the heat liberated of isothermal mixing at 300 K

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x4H4−x1H1−x2H2−x3H3

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Mass of 25-wt-% H₂SO₄= m₃ Total mass = m₄

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 25-wt-% H₂SO₄ = H₃ Total enthalpy = H₄

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Mass fraction of 25-wt-% H₂SO₄= x₃ Total mass fraction of H₂SO₄= x₄

(b)

Interpretation Introduction

Interpretation:

To find the temperature of the intermediate solution formed

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x3H3−x1H1−x2H2

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Total mass = m₃

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Total enthalpy = H₃

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Total mass fraction of H₂SO₄= x₃

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 11, Problem 11.30P

Interpretation Introduction

(a)

Interpretation:

To find the heat liberated of isothermal mixing at 300 K

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x4H4−x1H1−x2H2−x3H3

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Mass of 25-wt-% H₂SO₄= m₃ Total mass = m₄

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 25-wt-% H₂SO₄ = H₃ Total enthalpy = H₄

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Mass fraction of 25-wt-% H₂SO₄= x₃ Total mass fraction of H₂SO₄= x₄

(b)

Interpretation Introduction

Interpretation:

To find the temperature of the intermediate solution formed

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x3H3−x1H1−x2H2

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Total mass = m₃

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Total enthalpy = H₃

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Total mass fraction of H₂SO₄= x₃

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 11, Problem 11.30P

Interpretation Introduction

(a)

Interpretation:

To find the heat liberated of isothermal mixing at 300 K

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x4H4−x1H1−x2H2−x3H3

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Mass of 25-wt-% H₂SO₄= m₃ Total mass = m₄

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 25-wt-% H₂SO₄ = H₃ Total enthalpy = H₄

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Mass fraction of 25-wt-% H₂SO₄= x₃ Total mass fraction of H₂SO₄= x₄

(b)

Interpretation Introduction

Interpretation:

To find the temperature of the intermediate solution formed

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x3H3−x1H1−x2H2

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Total mass = m₃

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Total enthalpy = H₃

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Total mass fraction of H₂SO₄= x₃

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 11, Problem 11.30P

Interpretation Introduction

(a)

Interpretation:

To find the heat liberated of isothermal mixing at 300 K

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x4H4−x1H1−x2H2−x3H3

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Mass of 25-wt-% H₂SO₄= m₃ Total mass = m₄

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 25-wt-% H₂SO₄ = H₃ Total enthalpy = H₄

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Mass fraction of 25-wt-% H₂SO₄= x₃ Total mass fraction of H₂SO₄= x₄

(b)

Interpretation Introduction

Interpretation:

To find the temperature of the intermediate solution formed

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x3H3−x1H1−x2H2

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Total mass = m₃

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Total enthalpy = H₃

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Total mass fraction of H₂SO₄= x₃

Answer 6

Chapter 11, Problem 11.30P

Interpretation Introduction

(a)

Interpretation:

To find the heat liberated of isothermal mixing at 300 K

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x4H4−x1H1−x2H2−x3H3

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Mass of 25-wt-% H₂SO₄= m₃ Total mass = m₄

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 25-wt-% H₂SO₄ = H₃ Total enthalpy = H₄

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Mass fraction of 25-wt-% H₂SO₄= x₃ Total mass fraction of H₂SO₄= x₄

Answer 7

Chapter 11, Problem 11.30P

Interpretation Introduction

(a)

Interpretation:

To find the heat liberated of isothermal mixing at 300 K

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x4H4−x1H1−x2H2−x3H3

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Mass of 25-wt-% H₂SO₄= m₃ Total mass = m₄

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 25-wt-% H₂SO₄ = H₃ Total enthalpy = H₄

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Mass fraction of 25-wt-% H₂SO₄= x₃ Total mass fraction of H₂SO₄= x₄

Answer 8

(b)

Interpretation Introduction

Interpretation:

To find the temperature of the intermediate solution formed

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x3H3−x1H1−x2H2

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Total mass = m₃

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Total enthalpy = H₃

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Total mass fraction of H₂SO₄= x₃

Answer 9

(b)

Interpretation Introduction

Interpretation:

To find the temperature of the intermediate solution formed

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 25-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing, Q is as given below,

Q=x3H3−x1H1−x2H2

Where,

Mass of H₂O= m₁

Mass of H₂SO₄= m₂ Total mass = m₃

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Total enthalpy = H₃

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂ Total mass fraction of H₂SO₄= x₃

Answer 10

Expert Solution & Answer

Answer 11

Expert Solution & Answer

Answer 12

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 13

Ch. 11 - Prob. 11.1P Ch. 11 - Prob. 11.2P Ch. 11 - Prob. 11.3P Ch. 11 - Prob. 11.4P Ch. 11 - Prob. 11.5P Ch. 11 - Prob. 11.6P Ch. 11 - Prob. 11.7P Ch. 11 - Prob. 11.8P Ch. 11 - Prob. 11.9P Ch. 11 - Prob. 11.10P

Want to see the full answer?

Chapter 11 Solutions