Chapter 5, Problem 5.32QP

A 3.00-L flask containing 2.0 mol of O₂ and 1.0 mol of N₂ is in a room that is at 22.0°C.

1. a How much (what fraction) of the total pressure in the flask is due to the N₂?
2. b The flask is cooled and the pressure drops. What happens, if anything, to the mole fraction of the O₂ at the lower temperature?
3. c L of liquid water is introduced into the flask containing both gases. The pressure is then measured about 45 minutes later. Would you expect the measured pressure to be higher or lower?
4. d Given the information in this problem and the conditions in part c, would it be possible to calculate the pressure in the flask after the introduction of the water? If it is not possible with the given information, what further information would you need to accomplish this task?

Interpretation Introduction

Interpretation:

The amount of pressure due to N₂ gas present in a 3.00 L flask containing 2.0 mol of O₂ and 1.0 mol of N₂ should be measured.

Concept Introduction:

Ideal gas equation:

At a constant temperature (K) and pressure (P), the volume (v) occupied by the no of moles of any gas is known as ideal gas equation.

Ideal gas equation:

$\text{PV}\text{=}\text{nRT}$

And the SI units are

$\begin{array}{l}\text{T= Temperature }\left({\text{273}}^{\text{0}}\text{K}\right)\text{ = Kelvin}\\ \text{n = no of moles}\left(\text{1}\text{mole =}\text{6}{\text{.023×10}}^{\text{23}}\text{atoms}\right)\text{ = mole}\\ \text{V= Volume }\left(\text{22}\text{.4 L}\right)\text{ = cubic}\text{meter}\left({\text{m}}^{\text{3}}\right)\\ \text{P = Pressure }\left(\text{1}\text{atm}\right)\text{ = pascal(Pa)}\\ \text{R= universal gas constant }\left(\text{8}\text{.314 }\frac{\text{joule}}{\text{mole}\text{.kelvin}}\right)\text{ = }\frac{\text{joule}}{\text{mole}\text{.kelvin}}\end{array}$

Answer to Problem 5.32QP

The total pressure fraction due to N₂ gas is $\frac{\text{1}}{\text{3}}$.

Explanation of Solution

The flask contains 1.0 mol of N₂ gas out of 3.0 mol container, so the fraction of N₂ gas in the container is $\frac{\text{1}}{\text{3}}$.

Conclusion

The amount of pressure due to N₂ gas present in a 3.00 L flask containing 2.0 mol of O₂ and 1.0 mol of N₂ was measured.

Interpretation Introduction

Interpretation:

The change in mole fraction of O₂ at lower temperature in the flask should be explained

Concept Introduction:

Ideal gas equation:

At a constant temperature (K) and pressure (P), the volume (v) occupied by the no of moles of any gas is known as ideal gas equation.

Ideal gas equation:

$\text{PV}\text{=}\text{nRT}$

And the SI units are

$\begin{array}{l}\text{T= Temperature }\left({\text{273}}^{\text{0}}\text{K}\right)\text{ = Kelvin}\\ \text{n = no of moles}\left(\text{1}\text{mole =}\text{6}{\text{.023×10}}^{\text{23}}\text{atoms}\right)\text{ = mole}\\ \text{V= Volume }\left(\text{22}\text{.4 L}\right)\text{ = cubic}\text{meter}\left({\text{m}}^{\text{3}}\right)\\ \text{P = Pressure }\left(\text{1}\text{atm}\right)\text{ = pascal(Pa)}\\ \text{R= universal gas constant }\left(\text{8}\text{.314 }\frac{\text{joule}}{\text{mole}\text{.kelvin}}\right)\text{ = }\frac{\text{joule}}{\text{mole}\text{.kelvin}}\end{array}$

Answer to Problem 5.32QP

There will be no change in mole fraction of ${\text{O}}_{\text{2}}$.since mole fraction is not a function of temperature in ideal gas equation

Explanation of Solution

From the ideal gas equation, $\text{PV}\text{=}\text{nRT}$ we know that decrease in volume can decrease amount of pressure but there is no relation between mole fraction and temperature.

Conclusion

The change in mole fraction of O₂ at lower temperature in the flask was explained.

Interpretation Introduction

Interpretation:

The change in pressure when the 1.0 L liquid water added to the flask containing two gases should be explained.

Concept Introduction:

Ideal gas equation:

At a constant temperature (K) and pressure (P), the volume (v) occupied by the no of moles of any gas is known as ideal gas equation.

Ideal gas equation:

$\text{PV}\text{=}\text{nRT}$

And the SI units are

$\begin{array}{l}\text{T= Temperature }\left({\text{273}}^{\text{0}}\text{K}\right)\text{ = Kelvin}\\ \text{n = no of moles}\left(\text{1}\text{mole =}\text{6}{\text{.023×10}}^{\text{23}}\text{atoms}\right)\text{ = mole}\\ \text{V= Volume }\left(\text{22}\text{.4 L}\right)\text{ = cubic}\text{meter}\left({\text{m}}^{\text{3}}\right)\\ \text{P = Pressure }\left(\text{1}\text{atm}\right)\text{ = pascal(Pa)}\\ \text{R= universal gas constant }\left(\text{8}\text{.314 }\frac{\text{joule}}{\text{mole}\text{.kelvin}}\right)\text{ = }\frac{\text{joule}}{\text{mole}\text{.kelvin}}\end{array}$

Answer to Problem 5.32QP

The amount of pressure in the flask increases.

Explanation of Solution

According to ideal gas equation, the pressure in the flask is increases for two reasons

1. 1. When the water enters the flask occupies some spaces occupied by the gas earlier making gas molecules increase in pressure.
2. 2.  When the liquid water enters the flask it would evaporate after a time, and evaporate water would contribute the increase in pressure.

Conclusion

The change in pressure when the 1.0 L liquid water added to the flask containing two gases was explained.

Interpretation Introduction

Interpretation:

Using the parameters given in the part c is it possible to calculate the change in pressure should be explained.

Concept Introduction:

Ideal gas equation:

At a constant temperature (K) and pressure (P), the volume (v) occupied by the no of moles of any gas is known as ideal gas equation.

Ideal gas equation:

$\text{PV}\text{=}\text{nRT}$

And the SI units are

$\begin{array}{l}\text{T= Temperature }\left({\text{273}}^{\text{0}}\text{K}\right)\text{ = Kelvin}\\ \text{n = no of moles}\left(\text{1}\text{mole =}\text{6}{\text{.023×10}}^{\text{23}}\text{atoms}\right)\text{ = mole}\\ \text{V= Volume }\left(\text{22}\text{.4 L}\right)\text{ = cubic}\text{meter}\left({\text{m}}^{\text{3}}\right)\\ \text{P = Pressure }\left(\text{1}\text{atm}\right)\text{ = pascal(Pa)}\\ \text{R= universal gas constant }\left(\text{8}\text{.314 }\frac{\text{joule}}{\text{mole}\text{.kelvin}}\right)\text{ = }\frac{\text{joule}}{\text{mole}\text{.kelvin}}\end{array}$

Answer to Problem 5.32QP

Yes, the given information is sufficient to calculate the change in pressure.

Explanation of Solution

For calculating pressure from ideal gas equation, we have to know the volume and temperature of the given gas. As the volume of the flask is fixed and temperature is lowered and it can be measured and the gas constant “R’’ value we knew it is a constant and the value of water vapor pressure should be known so that we can calculate the pressure in the flask.

Conclusion

The change in pressure when the 1.0 L liquid water added to the flask containing two gases was explained.