(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

Students have asked these similar questions

please provide me the solution with more details. because the previous solution is not clear

please, provide me the solution with details.

please, provide me the solution with details

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