Chapter 9, Problem 1QAP

(a)

Interpretation Introduction

Interpretation:

The pressure of the vapors of methyl alcohol is to be calculated under the given conditions.

Concept Introduction :

The ideal gas equation is the expression that relates different measurable properties of a gas. The expression is given as:

  $PV=nRT$

where,

P = Pressure of the gas

V = Volume of the container in which the gas is occupied

R = Ideal gas constant = 0.0821 L atm/ mol K

n = number of moles of the gas = Mass of gas / Molecular mass of gas

T = Absolute temperature of the gas i.e., temperature on Kelvin scale

Answer to Problem 1QAP

The pressure of the vapor at 35 ⁰C is 237 mm Hg and at 45 ⁰C is 245 mm Hg.

Explanation of Solution

Given:

P₁ = 254 mm Hg

T₁ = 57 ⁰C = (57+273) K = 330 K

T₂ = 35 ⁰C = (35+273) K = 308 K and T₂ = 45 ⁰C = (45+273) K = 318 K

Considering the ideal gas equation for two different conditions of temperature and pressure,

  $\begin{array}{l}PV=nRT\\ \frac{P_1{V}_1}{nR{T}_1}=\frac{P_2{V}_2}{nR{T}_2}\end{array}$

Since, volume is constant so, the pressure of the vapor under second condition can be calculated as

  $\begin{array}{l}PV=nRT\\ \frac{P_1{V}_1}{nR{T}_1}=\frac{P_2{V}_2}{nR{T}_2}\end{array}$

For T₂ = 35 ⁰C = (35+273) K = 308 K:

  $\begin{array}{l}\frac{254\text{ mm Hg}}{330\text{ K}}=\frac{P_2}{308\text{ K}}\\ \Rightarrow P_2=237\text{ mm Hg}\end{array}$

For T₂ = 45 ⁰C = (45+273) K = 318 K:

  $\begin{array}{l}\frac{254\text{ mm Hg}}{330\text{ K}}=\frac{P_2}{318\text{ K}}\\ \Rightarrow P_2=245\text{ mm Hg}\end{array}$

Therefore, the pressure of the vapor at 35 ⁰C is 237 mm Hg and at 45 ⁰C is 245 mm Hg.

(b)

Interpretation Introduction

Interpretation:

The pressures calculated in part (a) are to be compared with the equilibrium pressures of methyl alcohol.

Concept Introduction :

If the pressure of vapor is more than the equilibrium pressure, it indicates that the liquid has undergone condensation.

Answer to Problem 1QAP

At 35 ⁰C, the vapor pressure is more than the equilibrium pressure and at45 ⁰C, the vapor pressure is less than the equilibrium pressure.

Explanation of Solution

Given:

Equilibrium vapor pressure of methyl alcohol at

35 ⁰C = 203 mm Hg

45 ⁰C = 325 mm Hg

The calculated pressure of methyl alcohol at

35 ⁰C = 237 mm Hg

45 ⁰C = 245 mm Hg

Thus, at 35 ⁰C, the equilibrium vapor pressure of methyl alcohol is more than the calculated pressure, which implies that some of the vapors condense. The vapor pressure at 45 ⁰C is less than the equilibrium vapor pressure, which implies the equilibrium stage has still not reached and the liquid vaporizes.

(c)

Interpretation Introduction

Interpretation:

The pressure of methyl alcohol vapors inside the flask is to be calculated at 35 ⁰C and 45 ⁰C.

Concept Introduction : When a liquid is present in a closed container, it starts form vapors above the liquid surface. A stage is reached when the liquid and vapors are present in equilibrium with each other.

Answer to Problem 1QAP

Pressure of methyl alcohol vapors

At 35 ⁰C = 203 mm Hg

At 45 ⁰C = 325 mm Hg

Explanation of Solution

Once equilibrium between a liquid and its vapors is reached in a closed container, the liquid keeps on changing to vapors and vapors keep on changing to liquid. No further increase in pressure exerted by the vapors is observed.

Since, at 35 ⁰C, the equilibrium pressure of methyl alcohol vapors is less than the calculated pressure, the pressure that can actually be achieved is the equilibrium pressure,i.e. 203 mm Hg.

At 45 ⁰C, the equilibrium pressure of methyl alcohol vapors is less than the calculated pressure, thus the liquids are still getting converted to reach equilibrium. The system will reach equilibrium to exert a pressure of 325 mm Hg.

(d)

Interpretation Introduction

Interpretation:

The physical state of methyl alcohol present in the flask at two different temperatures, 35 ^oC and 45 ^oC is to be calculated.

Concept Introduction:

Vaporization results in the formation of vapors from liquid and condensation results in the formation of liquids from vapors.

Answer to Problem 1QAP

At 35 ⁰C, the flask mainly contains liquid and at 45 ⁰C, the flask mainly contains vapors.

Explanation of Solution

As shown in the above parts, at 35 ⁰C, condensation is taking place and thus vapors are getting converted to liquid. Thus, the flask mainly contains liquid.

At 45 ⁰C, vaporization is taking place and thus more liquid molecules are getting converted to vapors and thus the flask mainly contains vapor state.

Therefore, at 35 ⁰C, the flask contains mainly liquid state and at 45 ⁰C, the flask mainly contains vapor state.