Chapter 5, Problem 86QAP

The following figure shows three 1.00-L bulbs connected by valves. Each bulb contains argon gas with amounts proportional to the number of circles pictorially represented in the chamber. All three bulbs are maintained at the same temperature. Unless stated otherwise, assume that the valves connecting the bulbs are closed and seal the gases in their respective chambers. Assume also that the volume between each bulb is negligible.

(a) Which bulb has the highest pressure?

(b) If the pressure in bulb A is 0.500 atm, what is the pressure in bulb C?

(c) If the pressure in bulb A is 0.500 atm, what is the total pressure?

(d) If the pressure in bulb A is 0.500 arm, and the valve between bulbs A and B is opened, redraw the figure shown above to accurately represent the gas atoms in all three bulbs. What is $P_{\text{A}}+P_{\text{B}}+P_{\text{C}}$? Compare your answer in part (d) to that in part (c).

(e) Follow the instructions of part (d) but now open only the valve between bulbs B and C.

Interpretation Introduction

(a)

Interpretation:

The bulb with the highest pressure needs to be identified based on the given description.

Concept introduction:

The ideal gas equation is a thermodynamic equation of state which relates the pressure (P), volume (V), number of moles (n) and temperature (T) of an ideal gas through the following expression:

$\text{PV = nRT -----(1)}$

where R is the universal gas constant = 0.0821 L.atm/mol-K

Answer to Problem 86QAP

Bulb C has the highest pressure.

Explanation of Solution

Given Information:

Volume (V) of bulb A = B = C = 1.0 L

Temperature (T) of bulb A = B = C

Number of moles (n) of Ar gas in A = 2

Number of moles (n) of Ar gas in B = 4

Number of moles (n) of Ar gas in C = 10

Calculation:

Based on equation (1), the pressure (P) in each of the bulbs can de deduced by substituting the values of n in bulbs A, B and C and volume, V = 1.0 L

$\begin{array}{l}{\text{P}}_{\text{A}}\text{ = 2RT}\\ {\text{P}}_{\text{B}}\text{ = 4RT}\\ {\text{P}}_{\text{C}}\text{ = 10RT}\end{array}$

Now, pressure (P) is directly proportional to the number of moles (n). Therefore, under constant temperature, bulb C will have the highest pressure.

Interpretation Introduction

(b)

Interpretation:

The pressure in bulb C needs to be deduced if the pressure in bulb A is 0.500 atm.

Concept introduction:

The ideal gas equation is a thermodynamic equation of state which relates the pressure (P), volume (V), number of moles (n) and temperature (T) of an ideal gas through the following expression:

$\text{PV = nRT -----(1)}$

where R is the universal gas constant = 0.0821 L.atm/mol-K

Answer to Problem 86QAP

Pressure in Bulb C is 2.5 atm

Explanation of Solution

Given Information:

Volume (V) of bulb A = B = C = 1.0 L

Temperature (T) of bulb A = B = C

Pressure (P) in bulb A = 0.500 atm

Number of moles (n) of Ar gas in A = 2

Number of moles (n) of Ar gas in C = 10

Calculation:

Based on equation (1), the pressure (P) in bulbs A and C can be deduced by substituting the given values of n, V and P under constant T

$\begin{array}{l}{\text{P}}_{\text{A}}\text{ = 2RT}\\ {\text{P}}_{\text{C}}\text{ = 10RT}\\ \text{Now, }\frac{{\text{P}}_{\text{C}}}{{\text{P}}_{\text{A}}}=\frac{\text{10RT}}{\text{2RT}}=5\\ {\text{P}}_{\text{C}}{\text{ = 5P}}_{\text{A}}\\ \text{i}\text{.e}{\text{. P}}_{\text{C}}\text{ = 5(0}\text{.500 atm) = 2}\text{.5 atm}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The total pressure needs to be deduced if the pressure in bulb A is 0.500 atm.

Concept introduction:

The ideal gas equation is a thermodynamic equation of state which relates the pressure (P), volume (V), number of moles (n) and temperature (T) of an ideal gas through the following expression:

$\text{PV = nRT -----(1)}$

where R is the universal gas constant = 0.0821 L.atm/mol-K

As per Dalton’s law, the total pressure exerted by a gas mixture is equal to the sum of the partial pressure of the individual gases.

${\text{P}}_{\text{total}}{\text{ = P}}_{\text{1}}{\text{ + P}}_{\text{2}}{\text{ + P}}_{\text{3}}\text{ +}\mathrm{...}\text{ ------(2)}$

Answer to Problem 86QAP

Total pressure = 4.00 atm

Explanation of Solution

Given Information:

Volume (V) of bulb A = B = C = 1.0 L

Temperature (T) of bulb A = B = C

Pressure (P) in bulb A = 0.500 atm

Number of moles (n) of Ar gas in A = 2

Number of moles (n) of Ar gas in B = 4

Number of moles (n) of Ar gas in C = 10

Calculation:

Based on equation (1), the pressure (P) in bulbs A and B can be deduced by substituting the given values of n, V and P under constant T

$\begin{array}{l}{\text{P}}_{\text{A}}\text{ = 2RT}\\ {\text{P}}_{\text{B}}\text{ = 4RT}\\ \text{Now, }\frac{{\text{P}}_{\text{B}}}{{\text{P}}_{\text{A}}}=\frac{\text{4RT}}{\text{2RT}}=2\\ {\text{P}}_{\text{B}}{\text{ = 2P}}_{\text{A}}\\ \text{i}\text{.e}{\text{. P}}_{\text{B}}\text{ = 2(0}\text{.500 atm) = 1}\text{.00 atm}\\ {\text{From part (a), we know that P}}_{\text{C}}\text{ = 2}\text{.50 atm}\\ {\text{Total pressure = P}}_{\text{A}}{\text{ + P}}_{\text{B }}{\text{+ P}}_{\text{C}}=0.500+1.00+2.50=4.00\text{ atm}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The total pressure needs to be deduced after the valve between A and B is opened.

Concept introduction:

The ideal gas equation is a thermodynamic equation of state which relates the pressure (P), volume (V), number of moles (n) and temperature (T) of an ideal gas through the following expression:

$\text{PV = nRT -----(1)}$

where R is the universal gas constant = 0.0821 L.atm/mol-K

As per Dalton’s law, the total pressure exerted by a gas mixture is equal to the sum of the partial pressure of the individual gases.

${\text{P}}_{\text{total}}{\text{ = P}}_{\text{1}}{\text{ + P}}_{\text{2}}{\text{ + P}}_{\text{3}}\text{ +}\mathrm{...}\text{ ------(2)}$

Answer to Problem 86QAP

Total pressure after the valve between A and B is opened in 8.50 atm

Explanation of Solution

Given Information:

Volume (V) of bulb A = B = C = 1.0 L

Temperature (T) of bulb A = B = C

Pressure (P) in bulb A = 0.500 atm

Number of moles (n) of Ar gas in A = 2

Number of moles (n) of Ar gas in B = 4

Number of moles (n) of Ar gas in C = 10

Calculation:

When the valve between A and B is opened the Ar gas will diffuse from the region of high pressure i.e. bulb B to A until equilibrium is established.

Now the total number of moles (atoms) of Ar = 2 + 4 = 6. The final pressure in each bulb will be due to 6 moles (atoms) of Ar

Step 1: Calculate the final pressure in bulb A after mixing:

The initial pressure in bulb A = P_i = 0.500 atm

Initial moles of Ar gas in A = n_i = 2

Final moles in A = n_f = 6

The final pressure in bulb A = P_f

Under constant V and T, the ratio of the initial and final pressures would be:

$\begin{array}{l}\frac{{\text{P}}_{\text{f}}}{{\text{P}}_{\text{i}}}=\frac{{\text{n}}_{\text{f}}\text{RT/V}}{{\text{n}}_{\text{i}}\text{RT/V}}=\frac{6}{2}=3\\ {\text{P}}_{\text{f}}{\text{ = 3P}}_{\text{i}}\\ \text{i}\text{.e}{\text{. P}}_{\text{f}}\text{ = 3(0}\text{.500 atm) = 1}\text{.50 atm}\end{array}$

Step 2: Calculate the final pressure in bulb B after mixing:

The initial pressure in bulb B = P_i = 1.50 atm

Initial moles of Ar gas in A = n_i = 2

Final moles in A = n_f = 6

The final pressure in bulb A = P_f

Under constant V and T, the ratio of the initial and final pressures would be:

$\begin{array}{l}\frac{{\text{P}}_{\text{f}}}{{\text{P}}_{\text{i}}}=\frac{{\text{n}}_{\text{f}}\text{RT/V}}{{\text{n}}_{\text{i}}\text{RT/V}}=\frac{6}{2}=3\\ {\text{P}}_{\text{f}}{\text{ = 3P}}_{\text{i}}\\ \text{i}\text{.e}{\text{. P}}_{\text{f}}\text{ = 3(1}\text{.50 atm) = 4}\text{.50 atm}\end{array}$

Step 3: Calculate the total pressure after mixing:

$\begin{array}{l}{\text{Total pressure = P}}_{\text{A}}{\text{ + P}}_{\text{B }}{\text{+ P}}_{\text{C}}\\ =1.50+4.50+2.50=8.50\text{ atm}\end{array}$

The total pressure after the valve between A and B is opened is higher than that when the valve is closed.

Interpretation Introduction

(d)

Interpretation:

The total pressure needs to be deduced after the valve between B and C is opened.

Concept introduction:

The ideal gas equation is a thermodynamic equation of state which relates the pressure (P), volume (V), number of moles (n) and temperature (T) of an ideal gas through the following expression:

$\text{PV = nRT -----(1)}$

where R is the universal gas constant = 0.0821 L.atm/mol-K

As per Dalton’s law, the total pressure exerted by a gas mixture is equal to the sum of the partial pressure of the individual gases.

${\text{P}}_{\text{total}}{\text{ = P}}_{\text{1}}{\text{ + P}}_{\text{2}}{\text{ + P}}_{\text{3}}\text{ +}\mathrm{...}\text{ ------(2)}$

Answer to Problem 86QAP

Total pressure after the valve between B and C is opened in 9.25 atm

Explanation of Solution

Given Information:

Volume (V) of bulb A = B = C = 1.0 L

Temperature (T) of bulb A = B = C

Pressure (P) in bulb A = 0.500 atm

Number of moles (n) of Ar gas in A = 2

Number of moles (n) of Ar gas in B = 4

Number of moles (n) of Ar gas in C = 10

Calculation:

When the valve between B and C is opened the Ar gas will diffuse from the region of high pressure i.e. bulb C to B until equilibrium is established.

Now the total number of moles (atoms) of Ar = 4 + 10 = 14. The final pressure in each bulb will be due to 14 moles (atoms) of Ar

Step 1: Calculate the final pressure in bulb B after mixing:

The initial pressure in bulb B = P_i = 1.50 atm

Initial moles of Ar gas in B = n_i = 4

Final moles in A = n_f = 10

The final pressure in bulb B = P_f

Under constant V and T, the ratio of the initial and final pressures would be:

$\begin{array}{l}\frac{{\text{P}}_{\text{f}}}{{\text{P}}_{\text{i}}}=\frac{{\text{n}}_{\text{f}}\text{RT/V}}{{\text{n}}_{\text{i}}\text{RT/V}}=\frac{14}{4}=3.5\\ {\text{P}}_{\text{f}}\text{ = 3}{\text{.5P}}_{\text{i}}\\ \text{i}\text{.e}{\text{. P}}_{\text{f}}\text{ = 3}\text{.5(1}\text{.50 atm) = 5}\text{.25 atm}\end{array}$

Step 2: Calculate the final pressure in bulb C after mixing:

The initial pressure in bulb C = P_i = 2.50 atm

Initial moles of Ar gas in C = n_i = 10

Final moles in C = n_f = 14

The final pressure in bulb A = P_f

Under constant V and T, the ratio of the initial and final pressures would be:

$\begin{array}{l}\frac{{\text{P}}_{\text{f}}}{{\text{P}}_{\text{i}}}=\frac{{\text{n}}_{\text{f}}\text{RT/V}}{{\text{n}}_{\text{i}}\text{RT/V}}=\frac{14}{10}=1.4\\ {\text{P}}_{\text{f}}\text{ = 1}{\text{.4P}}_{\text{i}}\\ \text{i}\text{.e}{\text{. P}}_{\text{f}}\text{ = 1}\text{.4(2}\text{.50 atm) = 3}\text{.50 atm}\end{array}$

Step 3: Calculate the total pressure after mixing:

$\begin{array}{l}{\text{Total pressure = P}}_{\text{A}}{\text{ + P}}_{\text{B }}{\text{+ P}}_{\text{C}}\\ =0.500+5.25+3.50=9.25\text{ atm}\end{array}$

The total pressure after the valve between B and C is opened is higher than that when the valve is closed.