Q. 1.31P - An Introduction to Thermal Physics

Question

Imagine some helium in a cylinder with an initial volume of 1 liter and an initial pressure of 1 atm. Somehow the helium is made to expand to a final volume of 3 liters, in such a way that its pressure rises in direct proportion to its volume.

1. Sketch a graph of pressure vs. volume for this process.
2. Calculate the work done on the gas during this process, assuming that there are no “other” types of work being done.
3. Calculate the change in the helium’s energy content during this process.
4. Calculate the amount of heat added to or removed from the helium during this process.
5. Describe what you might do to cause the pressure to rise as the helium expands.

Step-by-Step Solution

Answer

1. The graph between pressure and volume is a straight line but does not pass through the origin.
2. Work done on the gas is $\boxed{W=-405.2\text{\hspace{0.17em}J}}$.
3. Change in helium’s energy is $\boxed{\Delta U=1216.05\text{\hspace{0.17em}J}}$.
4. The amount of heat added is $\boxed{Q=1621.25\text{\hspace{0.17em}J}}$.
5. Pressure increases when kinetic energy increases. To increase pressure, kinetic should be increased and finally, it is concluded that one has to increase the temperature to increase pressure.

Part a. Step 1. Introduction.

The pressure versus volume diagram can be drawn by taking pressure on $y$-axis and volume on $x$-axis. In this process, since the pressure is directly proportional to the volume so the final pressure vs. volume graph is a straight line but does not pass through the origin.

Part a. Step 2. Explanation.

Since the pressure is directly proportional to the volume of helium gas so for this process,  the graph between pressure and volume is a straight line but does not pass through the origin. The required graph between pressure and volume is shown below:

Part a. Step 3. Conclusion.

Hence, the graph between pressure and volume is a straight line but does not pass through the origin.

Part b. Step 1. Given.

Initial volume is $V_i=1\text{\hspace{0.17em}liter}$

Final volume is $V_f=3\text{\hspace{0.17em}liter}$

Initial pressure is $P_i=1\text{\hspace{0.17em}atm}$

Part b. Step 2. Formula used.

Expression for work done on the system is

$W=-\underset{v_i}{\overset{v_f}{\int }}PdV$                          …… (1)

Here $P$ is the pressure, $V_i\text{\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}{V}_f$ are initial volume, and final volume.

Pressure is directly proportional to the volume as given i.e.

$P\propto V$             …… (2)

$P=VA$            …… (3)

Here $A$ is the proportionality constant.

Part b. Step 3. Calculation.

From equation (3) it can be simplified as for the initial and final case

$\frac{P_i}{V_i}=A$            …… (4)

And

$\frac{P_f}{V_f}=A$           …… (5)

From equations (4) and (5) 

$\frac{P_i}{V_i}=\frac{P_f}{V_f}$

$P_f=\frac{P_i{V}_f}{V_i}$                               …… (6)

Substitute $1\times {10}^{-3}$ for $V_i$ and $3\times {10}^{-3}$ for $V_f$ and $1.013\times {10}^5$ for $P_i$ in equation (6)

$\begin{array}{l}{P}_f=\frac{3\times {10}^{-3}\times 1.013\times {10}^5}{1\times {10}^{-3}}\\ P_f=3.04\times {10}^5\text{\hspace{0.17em}Pa}\end{array}$

In the graph between pressure and volume, the slope is calculated as

$A=\frac{P_f-P_i}{V_f-V_i}$                             …… (7)

 

Substitute $1\times {10}^{-3}$ for $V_i$ and $3\times {10}^{-3}$ for $V_f$, $1.013\times {10}^5$ for $P_i$ and $3.04\times {10}^5$ for $P_f$ in equation (7)

$\begin{array}{l}A=\frac{3.04\times {10}^5-1.013\times {10}^5}{3\times {10}^{-3}-{10}^{-3}}\\ A=1.013\times {10}^8{\text{\hspace{0.17em}Pa.m}}^{\text{-3}}\end{array}$

Hence the relation between pressure and volume is

$P=1.013\times {10}^{8}V$                                 ….. (8)

Substitute $1.013\times {10}^{8}V$ for $P$ in equation (1)

$W=-\underset{v_i}{\overset{v_f}{\int }}1.013\times {10}^{8}V.dV$                                 …… (9)

Again substitute the upper and lower limit of the above integral i.e. $1\times {10}^{-3}$ for $V_i$ and $3\times {10}^{-3}$ for $V_f$ in equation (9)

$\begin{array}{l}W=-\underset{{10}^{-3}}{\overset{3\times {10}^{-3}}{\int }}1.013\times {10}^{8}V.dV\\ W=-1.013\times {10}^8\underset{{10}^{-3}}{\overset{3\times {10}^{-3}}{\int }}V.dV\\ W=-1.013\times {10}^8{\left[\frac{V^2}{2}\right]}_{{10}^{-3}}^{3\times {10}^{-3}}\\ \boxed{W=-405.2\text{\hspace{0.17em}J}}\end{array}$

Part b. Step 4. Conclusion.

Hence, work done on the gas is $\boxed{W=-405.2\text{\hspace{0.17em}J}}$.

Part c. Step 1. Given.

Initial volume and final volume are $1\times {10}^{-3}$ and $3\times {10}^{-3}$ respectively and initial and final pressure are $1.013\times {10}^5$ and $3.04\times {10}^5$ respectively.

Part c. Step 2. Formula used.

Thermal energy can be expressed as

$U_{thermal}=Nf\left(\frac{1}{2}kT\right)$                            …… (10)

Here, $f$ stands for the degree of freedom, $N$ stands for a number of molecules, $k$ stands for Boltzmann constant, and $T$ stands for temperature in Kelvin.

The equation of ideal gas is

$PV=NkT$                              …… (11)

Here, $P$ is the pressure of the gas, $V$ is the volume of gas contained, $k$ is Boltzmann constant, $T$ is the temperature in Kelvin and $N$ is the number of molecules.

Part c. Step 3. Calculation.

Equation (10) can be simplified as by substitute $PV$ for $NkT$

$U_{thermal}=\frac{f}{2}PV$                                   …… (12)

Change in thermal energy due to expansion is

$\Delta U=P_f{V}_f-P_i{V}_i$                                …… (13)

Equation (12) can be written as a substitute $P_f{V}_f-P_i{V}_i$ for $\Delta U$

$\Delta U=\frac{3}{2}\Delta PV=\frac{3}{2}.\left(P_f{V}_f-P_i{V}_i\right)$                                  …… (14)

Substitute $1\times {10}^{-3}$ for $V_i$ and $3\times {10}^{-3}$ for $V_f$, $1.013\times {10}^5$ for $P_i$ and $3.04\times {10}^5$ for $P_f$ in equation (14)

$\begin{array}{l}\Delta U=\frac{3}{2}\left(3.04\times {10}^5\times 3\times {10}^{-3}-1.013\times {10}^5\times {10}^{-3}\right)\\ \boxed{\Delta U=1216.05\text{\hspace{0.17em}J}}\end{array}$

Part c. Step 4. Conclusion.

Hence, a change in helium’s energy is $\boxed{\Delta U=1216.05\text{\hspace{0.17em}J}}$.

Part d. Step 1. Given.

Change in energy is $\Delta U=1216.05\text{\hspace{0.17em}J}$ and work done is $W=-405.2\text{\hspace{0.17em}J}$.

Part d. Step 2. Formula used.

The equation for the first law of thermodynamics is
$Q=\Delta U-W$                           …… (15)

Where $Q$ is the amount of heat added or removed, $\Delta U$ is a change in internal energy, $W$ and is external work done

Part d. Step 3. Calculation.

Substitute $1216.05\text{\hspace{0.17em}J}$ for $\Delta U$ and $-405.2\text{\hspace{0.17em}J}$ for $W$ in equation (15)

$Q=\left(1216.05+405.2\right)\text{\hspace{0.17em}J}$

$\boxed{Q=1621.25\text{\hspace{0.17em}J}}$

Since the sign of $Q$ is positive so heat is added to the system.

Part d. Step 4. Conclusion.

Hence, the amount of heat added is $\boxed{Q=1621.25\text{\hspace{0.17em}J}}$.

Part e. Step 1. Introduction

The motion of a body is related to the change in kinetic energy of the body. The kinetic energy of the body is directly proportional to the change in temperature of the body. 

Part e. Step 2. Explanation.

Kinetic energy is directly proportional to the temperature. Similarly, pressure is also directly proportional to kinetic energy. So, it can be concluded that pressure is also proportional to temperature. Basically, to increase pressure, one has to increase temperature.

Part e. Step 3. Conclusion.

Hence, to increase pressure, one has to increase the temperature.