Question

An ideal monatomic gas is contained in a cylinder with a movable piston so that the gas can do work on the outside world, and heat can be added or removed as necessary. The figure shows various paths that the gas might take in expanding from an initial state whose pressure, volume, and temperature are $p_0$ , $V_0$ , and $T_0$ respectively. The gas expands to a state with final volume $4V_0$ . For some answers it will be convenient to generalize your results by using the variable $R_v = V_{\rm final}/V_{\rm initial}$ , which is the ratio of final to initial volumes (equal to 4 for the expansions shown in the figure.)

The figure shows several possible paths of the system in the pV plane. Although there are an infinite number of paths possible, several of those shown are special because one of their state variables remains constant during the expansion. These have the following names:

• Adiabiatic: No heat is added or removed during the expansion.
• Isobaric: The pressure remains constant during the expansion.
• Isothermal: The temperature remains constant during the expansion.

What happens to the temperature of the gas during an isobaric expansion?

Temperature increases. Temperature remains constant. Temperature decreases.

Part C

Which of the curves in the figure represents an isothermal process?

A B C D

Part D

Graphically, the work along any path in the pV plot ____________.

is the area to the left of the curve from $p_0$ to $p_final$ is the area under the curve from $V_0$ to $V_final$ requires knowledge of the temperature

Part E

Calculate $W_A$ , the work done by the gas as it expands along path A from $V_0$ to .

Express $W_A$ in terms of $p_0$ , $V_0$ , and $R_v$ .

W=

Part G

Which of the curves shown represents an adiabatic expansion?

A B C D

Answer 1

Concepts and reason

Use the concept of adiabatic, isobaric and isothermal processes to solve this problem.

Analyze the formula for isobaric process and determine how it depends on temperature to find the temperature dependence of isobaric process.

Analyze the condition of the isothermal process and then determine, which processes are isothermal.

Analyze the expression of work done to find which area represents the work done.

Fundamentals

An isothermal process is one in which the temperature of the gas remains the same throughout the process. The condition for isothermal process is given as follows,

$pV = K$

Here, $p$ is the pressure, $V$ is the volume and $K$ is a constant.

An isobaric process is one in which pressure remains the same throughout the process. The condition for an isobaric process is given as follows,

$\frac{V}{T} = K$

Here, $V$ and $T$ are the volume and temperature of gas respectively.

An adiabatic process is one in which no heat transfer occurs throughout the process. The condition for an adiabatic process is given as:

$p{V^\gamma } = K$

Here,$\gamma$ is the ratio of specific heats of the gas at constant pressure and constant volume respectively.

The work done for any thermodynamic process is given as:

$W = \int_{{V_i}}^{{V_f}} {pdV}$

Here, $W$ is total work done, $dV$ is a change in volume, ${V_f}$ is the final volume and ${V_i}$ is the initial volume.

(A)

For an isobaric process,

$\frac{V}{T} = K$

This means that the volume of the gas is directly proportional to its temperature. The temperature of the gas increases as the volume increases (volume increases while expansion) for an isobaric expansion.

(C)

The condition for isothermal expansion is given as:

$pV = K$

It can be written as follows.

${p_1}{V_1} = {p_2}{V_2}$

The condition hints that the graph of $p$ versus $V$ is a parabolic curve. So, it should be either B, C or D.

Check, which curve satisfies the condition of isothermal processes.

For paths B, C, and D, ${P_1}{V_1} = 1$

Now calculate ${p_2}{V_2}$ for each path.

For path B:

$\begin{array}{c}\\{P_2}{V_2} = \left( {0.625} \right)\left( 4 \right)\\\\ = 2.5\\\end{array}$

So, ${P_1}{V_1} \ne {P_2}{V_2}$

Hence, path B is not an isothermal process.

For path D,

$\begin{array}{c}\\{P_2}{V_2} = \left( {0.140} \right)\left( 4 \right)\\\\ = 0.56\\\end{array}$

Path D is not isothermal.

For path C:

$\begin{array}{c}\\{P_2}{V_2} = \left( {0.25} \right)\left( 4 \right)\\\\ = 1\\\end{array}$

So, ${P_1}{V_1} = {P_2}{V_2}$

Hence, path C is an isothermal process.

[Part C]

Part C

(D)

The work done for any thermodynamic process is given as follows,

$W = \int_{{V_i}}^{{V_f}} {pdV}$

The above equation is nothing but the area under the curve on $pV$ diagram projected on volume axis, within the limits ${V_i}$ and ${V_f}$.

(E)

Work done for an isobaric process is given as follows,

$W = p\left( {{V_2} - {V_1}} \right)$

Substitute, ${p_0}$ for $p$, ${R_v}{V_0}$ for ${V_2}$ and ${V_0}$ for ${V_1}$.

$\begin{array}{c}\\W = {p_0}\left( {{R_v}{V_0} - {V_0}} \right)\\\\ = {p_0}{V_0}\left( {{R_v} - 1} \right)\\\end{array}$

[Part E]

(G)

Condition for adiabatic processes is as follows,

${p_1}V_1^\gamma = {p_2}V_2^\gamma$

For a monoatomic gas $\gamma = 1.4$ hence, the above condition becomes as follows,

${p_1}V_1^{1.4} = {p_2}V_2^{1.4}$

For all processes,

$\begin{array}{c}\\{p_1}V_1^{1.4} = \left( 1 \right){\left( 1 \right)^{1.4}}\\\\ = 1\\\end{array}$

Now calculate ${p_2}V_2^{1.4}$ for each path.

For path A:

$\begin{array}{c}\\{P_2}V_2^{1.4} = \left( {1.00} \right){\left( 4 \right)^{1.4}}\\\\ = 6.96\\\end{array}$

For path B:

$\begin{array}{c}\\{P_2}V_2^{1.4} = \left( {0.625} \right){\left( 4 \right)^{1.4}}\\\\ = 4.35\\\end{array}$

For path C:

$\begin{array}{c}\\{P_2}V_2^{1.4} = \left( {0.25} \right){\left( 4 \right)^{1.4}}\\\\ = 1.74\\\end{array}$

For path D,

$\begin{array}{c}\\{P_2}V_2^{1.4} = \left( {0.140} \right){\left( 4 \right)^{1.4}}\\\\ = 0.975\\\end{array}$

Hence, only path D is an adiabatic expansion.

Ans: Part A

The temperature of the gas increases for an isobaric expansion.