Chapter 5, Problem 5.93P

To study a key fuel-cell reaction, a chemical engineer has 20.0-L tanks of ${\text{H}}_{\text{2}}$ and of ${\text{O}}_{\text{2}}$ and wants to use up both tanks to form 28.0 mol of water at 23.8°C. (a) Use the ideal gas law to find the pressure needed in each tank. (b) Use the van der Waals eqauation to find the pressure needed in each tank. (c) Compare the results from the two equations.

Interpretation Introduction

Interpretation:

The pressure required in each tank should be calculated by using ideal gas Law.

Concept Introduction:

Ideal gas law equation is shown as:

  $\text{PV= nRT}$

Where,

P = Pressure

V = Volume

n = Number of moles

R = Universal gas constant

T = Temperature

Answer to Problem 5.93P

Pressure needed in hydrogen tank is $\text{34}\text{.11 atm}$ .

Pressure needed in oxygen tank is $17.05\text{ atm}$ .

Explanation of Solution

Given information:

Volume of ${\text{H}}_{\text{2}}$ tank = 20.0 L

Volume of ${\text{O}}_{\text{2}}$ tank = 20.0 L

Number of moles of water in both tank 28.0 moles

Temperature = 23.8 $°\text{C}$

The chemical reaction between hydrogen and oxygen is:

  ${\text{2H}}_{\text{2}}+{\text{O}}_{\text{2}}\to 2{\text{H}}_{\text{2}}\text{O}$

According to the reaction, ratio between hydrogen and water is 1:1.

Thus, number of moles of hydrogen = 28.0 mole

According to the reaction, ratio between oxygen and water is 1:2.

Thus, number of moles of oxygen = $28.0\text{ mole }\times \frac{1}{2}=14.0\text{ mole}$

Temperature in K = $23.8+273\text{ = 296}\text{.8 K}$

The formula for calculating pressure of hydrogen tank is:

  $P=\frac{nRT}{V}$

Put the values,

  $P=\frac{28.0\text{ mol}\times 0.0821\text{ L}\cdot \text{atm/mol}\cdot \text{K}\times \text{296}\text{.8 K}}{20.0\text{ L}}$

   = $34.11\text{ atm}$

Thus, pressure needed in hydrogen tank is $34.11\text{ atm}$ .

The formula for calculating pressure of oxygen tank is:

  $P=\frac{nRT}{V}$

Put the values,

  $P=\frac{14.0\text{ mol}\times 0.0821\text{ L}\cdot \text{atm/mol}\cdot \text{K}\times \text{296}\text{.8 K}}{20.0\text{ L}}$

   = $17.05\text{ atm}$

Thus, pressure needed inoxygen tank is $17.05\text{ atm}$ .

Interpretation Introduction

Interpretation:

The pressure required in both tank should be calculated by using Van der Waals equation.

Concept Introduction:

Van der Waals equation is shown as:

  $\text{P= }\frac{\text{nRT}}{V-nb}-a{\left(\frac{n}{V}\right)}^2$

Where,

P = Pressure

V = Volume

n = Number of moles

R = Universal gas constant

T = Temperature

a and b are constants.

Answer to Problem 5.93P

Van der Waal’s Pressure ( $P+\frac{n^{2}a}{V^2}$ ) needed in hydrogen tank is $35.44\text{ atm}$ .

Van der Waal’s Pressure ( $P+\frac{n^{2}a}{V^2}$ ) needed in oxygen tank is $17.44\text{ atm}$ .

Explanation of Solution

For hydrogen tank:

Value of constant a = $0.244\text{ atm}\cdot {\text{L}}^2/{\text{mol}}^{\text{2}}$

Value of constant b = $0.0266\text{ L}/\text{mol}$

The formula for calculating pressure of hydrogen tank is:

  $\text{P= }\frac{\text{nRT}}{V-nb}-a{\left(\frac{n}{V}\right)}^2$

Put the values,

  $\text{P= }\frac{\text{28 mol}\times \text{0}\text{.0821 atmL/Kmol}\times 296.8\text{ K}}{20.0\text{ L}-28\text{ mol}\times 0.0266\text{ L/mol}}-0.244\text{ atm}\cdot {\text{L}}^2{\text{mol}}^{-2}{\left(\frac{28.0\text{ mol}}{20.0\text{ L}}\right)}^2$

  $\text{ = }\frac{682.3\text{ L}\cdot \text{atm}}{19.25\text{ L}}-\frac{191.296\text{ atm}\cdot {\text{L}}^2/{\text{mol}}^{\text{2}}\cdot {\text{mol}}^{\text{2}}}{400{\text{ L}}^2}$

  $\text{= }35.44\text{ atm}-0.478\text{ atm}$

  $\text{= 34}\text{.96 atm}$

Thus, pressure needed in hydrogen tank is $34.96\text{ atm}$ .

Van der Waal’s Pressure ( $P+\frac{n^{2}a}{V^2}$ ) needed in hydrogen tank is $35.44\text{ atm}$ .

For oxygen tank:

Value of constant a = $1.36\text{ atm}\cdot {\text{L}}^2/{\text{mol}}^{\text{2}}$

Value of constant b = $0.0318\text{ L}/\text{mol}$

The formula for calculating pressure of hydrogen tank is:

  $\text{P= }\frac{\text{nRT}}{V-nb}-a{\left(\frac{n}{V}\right)}^2$

Put the values,

  $\text{P= }\frac{\text{14}\text{.0 mol}\times \text{0}\text{.0821 atmL/Kmol}\times 296.8\text{ K}}{20.0\text{ L}-14.0\text{ mol}\times 0.0318\text{ L/mol}}-1.36\text{ atm}\cdot {\text{L}}^2{\text{mol}}^{-2}{\left(\frac{14.0\text{ mol}}{20.0\text{ L}}\right)}^2$

  $\text{ = }\frac{\text{341 L}\cdot \text{atm}}{19.55\text{ L}}-\frac{266.56\text{ atm}\cdot {\text{L}}^2/{\text{mol}}^{\text{2}}\cdot {\text{mol}}^{\text{2}}}{400{\text{ L}}^2}$

  $\text{= 17}\text{.44 atm}-0.6664\text{ atm}$

  $\text{= 16}\text{.77 atm}$

Thus, pressure needed in oxygen tank is $16.77\text{ atm}$

Van der Waal’s Pressure ( $P+\frac{n^{2}a}{V^2}$ ) needed in oxygen tank is $17.44\text{ atm}$ .

Interpretation Introduction

Interpretation:

The results or pressure of values from the two equations should be compared.

Concept Introduction:

Ideal gas law equation is shown as:

  $\text{PV= nRT}$

Where,

P = Pressure

V = Volume

n = Number of moles

R = Universal gas constant

T = Temperature

Van der Waals equation is shown as:

  $\text{P= }\frac{\text{nRT}}{V-nb}-a{\left(\frac{n}{V}\right)}^2$

Where,

P = Pressure

V = Volume

n = Number of moles

R = Universal gas constant

T = Temperature

a and b are constants.

Answer to Problem 5.93P

The Van der Waal’s pressures are higher.

Explanation of Solution

The Van der Waals pressures which are calculated with Van der Waals equation are higher than the pressure calculated from ideal gas law. The reason behind this is the pressure correction in Van der Waal’s equation. The attraction present between the molecules decreases the pressure or force of collision and Van der Waal’s equation adjust the pressure high.

Hence, the Van der Waal’s pressures are higher.