Chapter 5, Problem 5.31QP

(a)

Interpretation Introduction

Interpretation:

Considering a gas container equipped with a movable piston, the change in volume and 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 5.31QP

An increase of pressure by 2 times to the original, will decrease volume by $\frac{\text{1}}{\text{2}}\text{(C}\text{to}\text{D)}$ to its original.

Explanation of Solution

Figure 1

From Ideal gas law,

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

Pressure and volume are inversely proportional to each other. So the increase in pressure causes decrease in volume C to D.

Conclusion

The change in volume with respect to pressure was explained.

(b)

Interpretation Introduction

Interpretation:

Considering a gas container equipped with a movable piston, the change in volume and 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 5.31QP

An increase of volume (C to A) by 2 times to the original, will decrease pressure by $\frac{\text{1}}{\text{2}}$ to its original.

Explanation of Solution

Figure 1

From Ideal gas law,

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

Pressure and volume are inversely proportional to each other. So the increase in volume (C to A) causes decrease in pressure as molecules get more space to move around.

Conclusion

The change in volume with respect to pressure was explained.

(c)

Interpretation Introduction

Interpretation:

Considering a gas container equipped with a movable piston, the change in temperature when volume changes from C to B 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 5.31QP

The change in kelvin temperature by 1.5 factor would cause the change in volume (C to B)

Explanation of Solution

Figure 1

From Ideal gas law,

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

Temperature and volume are proportional to each other. So the increase in volume (C to A) causes increase in temperature.

Conclusion

The change in volume with respect to temperature was explained

(d)

Interpretation Introduction

Interpretation:

The change in volume and pressure when the number of moles increased in a container with a moving piston 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 5.31QP

The increase in the number of moles by 2 factor would cause the increase in volume by 2 factor (C to A).

Explanation of Solution

Figure 1

From Ideal gas law,

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

Number of particles and volume are proportional to each other. So the increase in molecules would increase the volume by 2 factor (C to A). Since the piston is movable the pressure will not be affected (similar to starting pressure) even though number of particles and pressure are directly proportional to each other.

Conclusion

The change in volume and pressure when the number of moles increased in a container with a moving piston was explained.