Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 5, Problem 5.125P

(a)

Interpretation Introduction

Interpretation:

The pressure of tanks of H2 and of O2 at 23.8 °C by the ideal gas equation is to be calculated.

Concept introduction:

According to Boyle’s law, the volume occupied by the gas is inversely proportional to the pressure at the constant temperature.

The relationship between pressure and volume can be expressed as follows,

PV=constant

Here, P is the pressure and V is the volume.

According to Charles's law, the volume occupied by the gas is directly proportional to the temperature at the constant pressure.

The relationship between pressure and temperature can be expressed as follows,

V α T

Here, T is the temperature and V is the volume.

According to Avogadro’s law, the volume occupied by the gas is directly proportional to the mole of the gas at the constant pressure and temperature.

The relationship between volume and mole can be expressed as follows,

α n

Here, n is the mole of the gas and V is the volume.

The ideal gas equation can be expressed as follows,

PV=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

(a)

Expert Solution
Check Mark

Answer to Problem 5.125P

The pressure of tanks by the ideal gas equation of H2 is 34.1 atm and O2 is 17.1 atm

Explanation of Solution

The equation for the reaction of H2 with O2 is as follows:

2H2(g)+ O2(g)2H2O(g) (1)

From the equation (1), two moles of the H2O will be formed one mole of O2 . Thus, the moles of O2 is calculated from H2O as follows:

Moles of O2=[(Moles of H2O)(1 mol O22 mol H2O)] (2)

Substitute the value 28 mol for moles of H2O in the equation (2).

Moles of O2=[(28 mol)(1 mol O22 mol H2O)]=14 mol

From the equation (1), two moles of the H2O will be formed two moles of H2 . Thus, the moles of H2 is calculated from H2O as follows:

Moles of H2=[(Moles of H2O)(2 mol H22 mol H2O)] (3)

Substitute the value 28 mol for moles of H2O in the equation (3).

Moles of H2=[(28 mol)(2 mol H22 mol H2O)]=28 mol

The formula to convert °C to Kelvin is:

TK=T°C+273.15 (4)

Substitute 23.8 °C for T°C in equation (4)

TK=23.8+273.15=296.95 K

The expression to calculate the pressure of H2 is as follows:

PIGLV=nRT (5)

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of H2 and R is the gas constant.

Rearrange the equation (5) to calculate PIGL as follows:

PIGL=nRTV (6)

Substitute the value 20.0 L for V, 296.15 K for T, 28 mol for n and 0.0821 LatmmolK for R in the equation (6).

PIGL=(28 mol)(0.0821 LatmmolK)(296.15 K)(20.0 L)=34.131 atm34.1 atm

The expression to calculate the moles of O2 is as follows:

PV=nRT (7)

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of O2 and R is the gas constant.

Rearrange the equation (7) to calculate n as follows:

n=PVRT (8)

Substitute the value 20.0 L for V, 296.15 K for T, 14 mol for n and 0.0821 LatmmolK for R in the equation (8).

PIGL=(14 mol)(0.0821 LatmmolK)(296.15 K)(20.0 L)=17.0657 atm17.1 atm

Conclusion

The pressure of tanks by the ideal gas equation of H2 is 34.1 atm and O2 is 17.1 atm.

(b)

Interpretation Introduction

Interpretation:

The pressure of tanks of H2 and of O2 at 23.8 °C by van der Waals equation is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows,

PV=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

The expression of the van der Waals equation is as follows:

(PVDW+n2aV2)(Vnb)=nRT (8)

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas, R is the gas constant, a and b are van der Waals constants.

(b)

Expert Solution
Check Mark

Answer to Problem 5.125P

The pressure of tanks by van der Waals equation of H2 is 35.0 atm and O2 is 16.8 atm.

Explanation of Solution

Rearrange the equation (8) to calculate the pressure of (H2)PVDW

PVDW=nRT(Vnb)n2aV2 (9)

Substitute the value 20 L for V, 296.95 K for T, 28 mol for n , 0.244 atmL2mol2 for a, 0.0266 Lmol for b and 0.0821 LatmmolK for R in the equation (9).

PVDW=[(28 mol)(0.0821 LatmmolK)(296.15 K)(20 L(28 mol)(0.0266 Lmol))(28 mol)2(0.244 atmL2mol2)(20 L)2]=34.978 atm35 atm

Rearrange the equation (8) to calculate the pressure of (O2)PVDW

PVDW=nRT(Vnb)n2aV2 (10)

Substitute the value 20 L for V, 296.95 K for T, 14 mol for n , 0.244 atmL2mol2 for a, 0.0266 Lmol for b and 0.0821 LatmmolK for R in the equation (10).

PVDW=[(14 mol)(0.0821 LatmmolK)(296.15 K)(20 L(14 mol)(0.0266 Lmol))(14 mol)2(0.244 atmL2mol2)(20 L)2]=16.7878 atm16.8 atm

Conclusion

The pressure of tanks by van der Waals equation of H2 is 35.0 atm and O2 is 16.8 atm.

(c)

Interpretation Introduction

Interpretation:

The result from the two equation is to be compared.

Concept introduction:

The ideal gas equation can be expressed as follows,

PV=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

The expression of the van der Waals equation is as follows:

(PVDW+n2aV2)(Vnb)=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas, R is the gas constant, a and b are van der Waals constants.

(c)

Expert Solution
Check Mark

Answer to Problem 5.125P

The van der Waals value for hydrogen has a higher value as compared to the ideal gas law but the van der Waals value has a lower value as compared to the ideal gas law.

Explanation of Solution

For hydrogen, the van der Waals value for hydrogen has a higher value as compared to the ideal gas law. For oxygen, the van der Waals value has a lower value as compared to the ideal gas law.

Conclusion

The van der Waals value for hydrogen has a higher value as compared to the ideal gas law but the van der Waals value has a lower value as compared to the ideal gas law.

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Chapter 5 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

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