Pressure exerted by 1.00 mol CO 2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation 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. The formula to calculate pressure of one mole of gas is as follows: P = R T ( V − b ) − a V 2 (2) 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.

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Answer 1

Question

Chapter 3, Problem 3E.5E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation 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.

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

P=RT(V−b)−aV2 (2)

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.

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 15.0 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)15.0 L=1.63 atm

Substitute 15.0 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(15.0 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(15.0 L)2)=1.62 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation is 1.63 atm and 1.62 atm respectively.

(b)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 0.500 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)0.500 L=48.9 atm

Substitute 0.500 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(0.500 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(0.500 L)2)=38.9 atm≈39 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation is 48.9 atm and 39 atm respectively.

(c)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 50 mL with help of ideal gas equation and van der Waals equation has to be determined. Also, reliability of ideal gas with pressure has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 50 mL for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(50 mL)(10−3 L1 mL)=489 atm

Substitute 50 mL for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)((50 mL)(10−3 L1 mL)−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2((50 mL)(10−3 L1 mL))2)=1.98×103 atm≈2×103 atm

Ideal gas equation and van der Waals equation have same values at low pressures but high differences arise when pressure is high.

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Answer 2

Question

Chapter 3, Problem 3E.5E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation 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.

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

P=RT(V−b)−aV2 (2)

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.

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 15.0 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)15.0 L=1.63 atm

Substitute 15.0 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(15.0 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(15.0 L)2)=1.62 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation is 1.63 atm and 1.62 atm respectively.

(b)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 0.500 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)0.500 L=48.9 atm

Substitute 0.500 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(0.500 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(0.500 L)2)=38.9 atm≈39 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation is 48.9 atm and 39 atm respectively.

(c)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 50 mL with help of ideal gas equation and van der Waals equation has to be determined. Also, reliability of ideal gas with pressure has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 50 mL for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(50 mL)(10−3 L1 mL)=489 atm

Substitute 50 mL for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)((50 mL)(10−3 L1 mL)−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2((50 mL)(10−3 L1 mL))2)=1.98×103 atm≈2×103 atm

Ideal gas equation and van der Waals equation have same values at low pressures but high differences arise when pressure is high.

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Answer 3

Question

Chapter 3, Problem 3E.5E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation 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.

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

P=RT(V−b)−aV2 (2)

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.

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 15.0 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)15.0 L=1.63 atm

Substitute 15.0 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(15.0 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(15.0 L)2)=1.62 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation is 1.63 atm and 1.62 atm respectively.

(b)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 0.500 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)0.500 L=48.9 atm

Substitute 0.500 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(0.500 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(0.500 L)2)=38.9 atm≈39 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation is 48.9 atm and 39 atm respectively.

(c)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 50 mL with help of ideal gas equation and van der Waals equation has to be determined. Also, reliability of ideal gas with pressure has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 50 mL for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(50 mL)(10−3 L1 mL)=489 atm

Substitute 50 mL for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)((50 mL)(10−3 L1 mL)−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2((50 mL)(10−3 L1 mL))2)=1.98×103 atm≈2×103 atm

Ideal gas equation and van der Waals equation have same values at low pressures but high differences arise when pressure is high.

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Answer 4

Question

Chapter 3, Problem 3E.5E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation 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.

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

P=RT(V−b)−aV2 (2)

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.

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 15.0 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)15.0 L=1.63 atm

Substitute 15.0 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(15.0 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(15.0 L)2)=1.62 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation is 1.63 atm and 1.62 atm respectively.

(b)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 0.500 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)0.500 L=48.9 atm

Substitute 0.500 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(0.500 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(0.500 L)2)=38.9 atm≈39 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation is 48.9 atm and 39 atm respectively.

(c)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 50 mL with help of ideal gas equation and van der Waals equation has to be determined. Also, reliability of ideal gas with pressure has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 50 mL for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(50 mL)(10−3 L1 mL)=489 atm

Substitute 50 mL for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)((50 mL)(10−3 L1 mL)−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2((50 mL)(10−3 L1 mL))2)=1.98×103 atm≈2×103 atm

Ideal gas equation and van der Waals equation have same values at low pressures but high differences arise when pressure is high.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 3, Problem 3E.5E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation 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.

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

P=RT(V−b)−aV2 (2)

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.

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 15.0 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)15.0 L=1.63 atm

Substitute 15.0 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(15.0 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(15.0 L)2)=1.62 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation is 1.63 atm and 1.62 atm respectively.

(b)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 0.500 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)0.500 L=48.9 atm

Substitute 0.500 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(0.500 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(0.500 L)2)=38.9 atm≈39 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation is 48.9 atm and 39 atm respectively.

(c)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 50 mL with help of ideal gas equation and van der Waals equation has to be determined. Also, reliability of ideal gas with pressure has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 50 mL for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(50 mL)(10−3 L1 mL)=489 atm

Substitute 50 mL for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)((50 mL)(10−3 L1 mL)−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2((50 mL)(10−3 L1 mL))2)=1.98×103 atm≈2×103 atm

Ideal gas equation and van der Waals equation have same values at low pressures but high differences arise when pressure is high.

Answer 6

Chapter 3, Problem 3E.5E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation 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.

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

P=RT(V−b)−aV2 (2)

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.

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 15.0 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)15.0 L=1.63 atm

Substitute 15.0 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(15.0 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(15.0 L)2)=1.62 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation is 1.63 atm and 1.62 atm respectively.

Answer 7

Chapter 3, Problem 3E.5E

(a)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation 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.

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

P=RT(V−b)−aV2 (2)

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 8

(a)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 15.0 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)15.0 L=1.63 atm

Substitute 15.0 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(15.0 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(15.0 L)2)=1.62 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation is 1.63 atm and 1.62 atm respectively.

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 CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 0.500 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)0.500 L=48.9 atm

Substitute 0.500 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(0.500 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(0.500 L)2)=38.9 atm≈39 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation is 48.9 atm and 39 atm respectively.

Answer 13

(b)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation has to be determined.

Concept Introduction:

Refer to part (a).

Answer 14

(b)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 0.500 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)0.500 L=48.9 atm

Substitute 0.500 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(0.500 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(0.500 L)2)=38.9 atm≈39 atm

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation is 48.9 atm and 39 atm respectively.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 50 mL with help of ideal gas equation and van der Waals equation has to be determined. Also, reliability of ideal gas with pressure has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 50 mL for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(50 mL)(10−3 L1 mL)=489 atm

Substitute 50 mL for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)((50 mL)(10−3 L1 mL)−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2((50 mL)(10−3 L1 mL))2)=1.98×103 atm≈2×103 atm

Ideal gas equation and van der Waals equation have same values at low pressures but high differences arise when pressure is high.

Answer 19

(c)

Interpretation Introduction

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 50 mL with help of ideal gas equation and van der Waals equation has to be determined. Also, reliability of ideal gas with pressure has to be determined.

Concept Introduction:

Refer to part (a).

Answer 20

(c)

Expert Solution

Explanation of Solution

Rearrange equation (1) to calculate P.

P=RTV (3)

Substitute 50 mL for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).

P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(50 mL)(10−3 L1 mL)=489 atm

Substitute 50 mL for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).

P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)((50 mL)(10−3 L1 mL)−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2((50 mL)(10−3 L1 mL))2)=1.98×103 atm≈2×103 atm

Ideal gas equation and van der Waals equation have same values at low pressures but high differences arise when pressure is high.

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

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

Concept explainers

Videos

Explanation of Solution

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 50 mL with help of ideal gas equation and van der Waals equation has to be determined. Also, reliability of ideal gas with pressure has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

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Chapter 3 Solutions

Concept explainers

Videos

Explanation of Solution

Interpretation: Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation has to be determined. Concept Introduction: Refer to part (a).

Explanation of Solution

Interpretation: Pressure exerted by 1.00 mol CO2 at 298 K and 50 mL with help of ideal gas equation and van der Waals equation has to be determined. Also, reliability of ideal gas with pressure has to be determined. Concept Introduction: Refer to part (a).

Explanation of Solution

Want to see more full solutions like this?

Chapter 3 Solutions

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation has to be determined.

Concept Introduction:

Refer to part (a).

Interpretation:

Pressure exerted by 1.00 mol CO2 at 298 K and 50 mL with help of ideal gas equation and van der Waals equation has to be determined. Also, reliability of ideal gas with pressure has to be determined.

Concept Introduction:

Refer to part (a).