Isothermal and Adiabatic Processes

By Jitender Singh on Feb 11, 2020

Properties of an Isothermal Process

The temperature T of the system is constant.
The equation of state is $pV$ = constant.
The work done by n moles of an ideal gas in an isothermal expansion from volume V₁ to V₂ at temperature T is given by \begin{align} W=nRT\ln\frac{V_2}{V_1} \end{align}
The change in internal energy is zero i.e., $\Delta U=0$.
The heat supplied to the gas is equal to the work done by the gas.

Properties of an Adiabatic Process

The heat supplied to (or from) the system is zero i.e., $\Delta Q=0$.
The equation of state is $pV^\gamma$ = constant.
The work done by n moles of an ideal gas in an adiabatic expansion from volume V₁ to V₂ at temperature T is given by \begin{align} W=\frac{p_1V_1-p_2V_2}{\gamma-1} \end{align}
By first law of thermodynamics, $\Delta U=-\Delta W$. The temperature of the gas decreases (i.e., $\Delta U < 0$) if work is done by the gas.

Solved Problems from IIT JEE

Problem from IIT JEE 2011

5.6 litre of helium gas at STP is adiabatically compressed to 0.7 litre. Taking the initial temperature to be $T_1$, the work done in the process is,

$\frac{9}{8}RT_{1}$
$\frac{3}{2}RT_{1}$
$\frac{15}{8}RT_{1}$
$\frac{9}{2}RT_{1}$

Solution: For an ideal gas, $pV=nRT$. In an adiabatic process, $TV^{\gamma-1}=\text{constant}$, and the work done by the gas is given by, \begin{alignat}{2} W&=\frac{p_1 V_1 -p_2 V_2}{\gamma -1} \\ &=\frac{nRT_1 -nRT_2}{\gamma -1} \\ &=\frac{nRT_1}{\gamma -1}\left[1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}\right], \nonumber \end{alignat} where $V_1=5.6\;\mathrm{litre}$ and $V_2=0.7\;\mathrm{litre}$ are initial and final volume of the gas. Now, 5.6 litre at STP gives $n=5.6/22.4=1/4\;\mathrm{mol}$. Helium, being a monoatomic gas, has $\gamma=5/3$. Substitute these values to get $W=-\frac{9}{8}RT_1$. Negative sign indicates that work is done on the gas.

Problem from IIT JEE 2010

Starting with the same initial conditions, an ideal gas expands from volume $V_1$ to $V_2$ in three different ways. The work done by the gas is $W_1$ if the process is purely isothermal, $W_2$ if purely isobaric and $W_3$ if purely adiabatic, then,

$W_2 > W_1 > W_3$
$W_2 > W_3 > W_1$
$W_1 > W_2 > W_3$
$W_1 > W_3 > W_2$

Solution: Let $p_1$, $V_1$, and $T_1$ be initial pressure, volume, and temperature of the gas. For an ideal gas, $p_1V_1=nRT_1$. The work done by the gas in isothermal, isobaric, and adiabatic processes are given by, \begin{align} W_1&=p_1V_1\ln \left(\frac{V_2}{V_1}\right), \nonumber\\ W_2&=p_1(V_2-V_1), \nonumber\\ W_3&=\frac{p_1V_1-p_2V_2}{\gamma-1} \\ &=\frac{p_1V_1}{\gamma-1}\left[1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}\right], \nonumber \end{align} where $p_2$ and $V_2$ are final pressure and volume. If relationship among $W_1$, $W_2$, and $W_3$ holds in general then it should hold good for any special case, say $V_2=2V_1$ and $\gamma=2$. In this case, $W_1=0.693 p_1 V_1$, $W_2=p_1 V_1$, and $W_3=0.5 p_1 V_1$, giving $W_2 > W_1 > W_3$.

Aliter: The isothermal, isobaric, and adiabatic processes starting at the same initial point ($p_1,V_1$) and having same final volume $V_2$ are drawn in the figure. The work done in a process is equal to the area under the curve on $p\text{-}V$ diagram. It is clear from the diagram that $W_2 > W_1 > W_3$.

Problem from IIT JEE 2004

An ideal gas expands isothermally from a volume $V_1$ to $V_2$ and then compressed to original volume $V_1$ adiabatically. Initial pressure is $p_1$ and final pressure is $p_3$. The total work done is $W$. Then,

$p_3 > p_1$, $W > 0$
$p_3 < p_1$, $W < 0$
$p_3 > p_1$, $W < 0$
$p_3 = p_1$, $W = 0$

Solution: On a $p\text{-}V$ diagram, magnitude of slope of an adiabatic process is greater than that of an isothermal process (adiabatic process is more steeper). This fact is applied at the point B to draw the isothermal process AB and the adiabatic process BC (see figure).

Note that slope is negative in both cases because it is defined as tangent ($\tan$) of an angle from $x$-axis measured in anti-clockwise direction. From the graph $p_3>p_1$. The work done by the gas is area under the $p\text{-}V$ curve taken positive for expansion and negative for contraction. Thus, $W_\text{AB}$ is positive and $W_\text{BC}$ is negative and $|W_\text{BC}|>|W_\text{AB}|$. Thus, total work done $W=W_\text{AB}+W_\text{BC}<0$. The readers are encouraged to find expressions for $p_3$ and $W$ in terms of given parameters. Hint: $p_3=p_1\left(V_2/V_1\right)^{\gamma-1}$, $W_\text{AB}=p_1V_1\ln\frac{V_2}{V_1}$, and $W_\text{BC}=\frac{p_1V_1}{\gamma-1}\left(1-(V_2/V_1)^{\gamma-1}\right)$.

Questions on Thermodynamic Process

Question 1: Which of the following is not true for an isobaric process?

Question 2: Which of the following is not true for an isochoric process?

Question 3: An isothermal and an adiabatic processes are shown on a p-V diagram. These processes intersect at a point A. The magnitude of the slope of adiabatic process at A is more than the magnitude of slope of isothermal process at A.

First law of thermodynamics and its applications