First Law of Thermodynamics
2.0 Three important terms in first law of thermodynamics.
2.0 Three important terms in first law of thermodynamics.
Zeroth law of thermodynamics:
If two bodies $A$ and $B$ are in thermal equilibrium, $B$ and $C$ are in thermal equilibrium, then $A$ and $C$ are also in thermal equilibrium.
Heat transfer ($Q$ or $dQ$):
A gas sample exchanges energy with surroundings by two quantities
(a) Heat
(b) Work
If heat is given to the system, it is represented by $$Q = mS\Delta T$$ where $m$ is the mass of the gas sample, $S$ is the specific heat, $\Delta T$ is the increase or decrease in temperature.
If it is studied in terms of moles, the specific heat $S$, changes to molar specific heat $C$.$$dQ = nCdT$$Where $n$ is number of moles, $C$ is molar specific heat.
If we integrate the above equation between the limits ${T_1}$ and ${T_2}$, we get
$$Q = \int\limits_{{T_1}}^{{T_2}} {nCdT} $$
If number of moles and molar specific heat of the gas are constant, we get $$Q = nC\Delta T$$ Where $\Delta T = {T_2} - {T_1}$
How to find molar specific heat of a gas?
The molar specific heat of gas is given for different process, derivation and explanation is given in further section.
Let us consider a process $P{V^x}$=constant, the $C$ can be written as $$C = \frac{R}{{\gamma - 1}} + \frac{R}{{1 - x}} = {c_v} + \frac{R}{{1 - x}}$$
For Isochoric process $$c = {c_v} = \frac{R}{{\gamma - 1}}$$
For Isobaric process $$c = {c_p} = {c_v} + R$$
Heat is the path function, i.e. it varies for different paths.
The work done in the path $1$ ,path $2$ are different as work done is path dependent.
Internal energy:
Internal energy of a system is the total of all kinds of energy possessed by the system due to the molecular motion and molecular configuration. It depends only on the temperature difference between the initial and the final states but not on the process between the states.
When heat is given to the system, gas may expand and do work on the surroundings. It may be possible that a part of heat increases the kinetic energy of the gas.
Now, we will discuss about internal energy of the gas.
Internal energy of a ideal gas is the total kinetic energy of the gas molecules. It is represented by$$U = n{C_V}T$$$n$=number of moles of the gas
${C_V}$=molar specific heat at constant volume
$T$=temperature
We can not find the internal energy of the gas, we can only find the change in internal energy.
Change in internal energy of the gas is given by,
$$\Delta U = n{C_V}\Delta T$$
In differential form it is given by,
$$dU = n{C_V}dT$$
Internal energy is a state function i.e. it depends only on the intial and final state or independent of path.
As internal energy is a state function, we can use the expression of internal energy in any type of process, like isothermal, adiabatic, isochoric process ....etc. The change in internal energy in different paths but with same initial and final states are equal.
Question 1. One mole of an ideal monatonic gas is taken at a temperature of $300\ K$. Its volume is doubled keeping its pressure constant. Find the change in internal energy.
Solution: Using the gas law,
Since the pressure is constant, therefore$$\begin{equation} \begin{aligned} V \propto T\ \ \ \Rightarrow\ \ \ \frac{{{V_i}}}{{{T_i}}} = \frac{{{V_f}}}{{{T_f}}} \\ \therefore {T_f} = \frac{{{V_f}}}{{{V_i}}}{T_i}\ \ \Rightarrow\ \ \ 2{T_i} = 600K \\ \therefore \Delta U = \frac{f}{2}nR\Delta T = \frac{3}{2}1R(600 - 300) = 450R \\\end{aligned} \end{equation} $$
Work done:
A container fitted with a movable piston of cross sectional area $A$ is an example of thermodynamic system shown in the figure. If the gas expands against the piston, it exerts a force through a distance and does work on the piston. If the piston compresses the gas as it is moved inward, work is also done , but in this case on the gas. If the gas pressure on the piston face is $P$. Then the force on the piston due to gas is $PA$.
When the piston is moved outward an infinitesimal distance $dx$, the work done by the gas is $$dW = F.dx = PAdx$$
The change in volume of the gas is $dV = Adx$, then the work done becomes $$dW = PdV$$
For a finite change in volume from ${V_i}$ to ${V_f}$, this equation is then integrated between ${V_i}$ to ${V_f}$ to find the net work $$W = \int {dW = \int\limits_{{V_i}}^{{V_f}} {PdV} } $$ Now there are two methods of finding work done by the gas.
The work is also a path function.
So, the work done in different paths is different. From the figure, the work done in path $1$, path $2$, path $3$ are different.
Method 1:
This is used when $P-V$ equation is known to us.If $P$ is a function of $V$, $P = f(V)$. Then the work done can be found by,$${W = \int\limits_{{v_i}}^{{v_f}} {f(V)dV} }$$ Where ${V_i}$ and ${V_f}$ are the initial and final volumes of the gas
Question 2. In any thermodynamic process, pressure ( in $Pa$) varies as $P = aV + b$ [ $P$ represents pressure, $V$ represents volume ]. Find work done by gas during expansion from $1\ {m^3}\ to\ 2\ {m^3}$.
Solution: $$\begin{equation} \begin{aligned} W = \int\limits_1^2 {\left( {aV + b} \right)} dV = \int\limits_1^2 {aVdV} + \int\limits_1^2 {bdV} \\ = \frac{a}{2}\left( {{2^2} - {1^2}} \right) + b\left( {2 - 1} \right) \\ = \frac{a}{2}\left( 3 \right) + b = \left( {\frac{{3a}}{2} + b} \right)J \\\end{aligned} \end{equation} $$
Method 2:
Case 1:
Volume ($V$) is constant:
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| As, $V$ is constant, And the area under the graph is zero. So, work done $(W_{AB})$ is zero. | As, $V$ is constant, And the area under the graph is zero. So, work done $(W_{BA})$ is zero. |
Case 2:
Volume is increasing:
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| Here the volume is increasing, simultaneously pressure is also increasing. Area under the graph represents the work done Here, $V$ is increasing, so ${W_{AB}} > 0$ ${W_{AB}}$ = Shaded region | Here the volume is increasing, simultaneously pressure is decreasing. Area under the graph represents the work done Here, $V$ is increasing, so ${W_{AB}} > 0$ ${W_{AB}}$ = Shaded region |
Case 3:
Volume is decreasing:
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Here the volume is decreasing, simultaneously pressure is also increasing. Area under the graph represents the work done Here, $V$ is decreasing, so ${W_{AB}} < 0$ ${W_{AB}}$ = -Shaded region | Here the volume is decreasing, simultaneously pressure is also decreasing. Area under the graph represents the work done Here, $V$ is decreasing, so ${W_{AB}} < 0$ ${W_{AB}}$ = -Shaded region |
Case 4:
Cyclic process:
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The work done is represented by the area enclosed by the gas. As the cycle is in clockwise direction, ${W_{clockwi\operatorname{se} \_cycle}}$= +shaded area | The work done is represented by the area enclosed by the gas. As the cycle is in clockwise direction, ${W_{anticlockwi\operatorname{se} \_cycle}}$= +shaded area |
Case 5:
Incomplete cycle:
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The work done is given by the shaded region ${W_{ABC}}$ = + shaded area | The work done is given by the shaded region ${W_{ABCD}}$ = - shaded area |
Question 3. Find the work done by the gas in the process shown below by $P$-$V$ diagram.
Solution:
Work done = Area bounded by $P$ to $V$ $$\begin{equation} \begin{aligned} = 200kPa \times 75\ c.c + \frac{1}{2} \times \left( {300} \right) \times kPa \times 25\ c.c \\ = \left( {200 \times 75 \times {{10}^3} \times {{10}^{ - 6}} + 150 \times {{10}^3} \times 25 \times {{10}^{ - 6}}} \right)J \\ = \left( {2 \times 75 \times {{10}^{ - 1}} + 15 \times 25 \times {{10}^{ - 2}}} \right)J \\ = 15 + 3.75 \\ = 18.75J \\\end{aligned} \end{equation} $$
