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1800-102-2727Heat is involved everywhere. Be it burning of tyres, cooking food or heating a gas in a cylinder, heat finds applications in many places. You might have come across the fact heat is a form of energy. In physics, energy is always capable of being converted to work. For instance, if you heat a pan containing water, the molecules absorb the heat and start spreading out fast. The heat energy supplied by the stove gets converted into kinetic energy of the molecules. Similarly, when a driver presses the pedals of brakes of the car, the kinetic energy of the wheels gets converted into heat energy. The first law of thermodynamics aims to explain this energy conservation. In this article, we will explore the first law of thermodynamics in detail.
Table of contents
Let us consider an insulated cylinder fitted with a piston on top. The cylinder is filled with gas. When heated, the gas expands, and tries to push the piston up. In this process, some amount of work is done by the gas. A portion of the heat supplied is used to increase the internal energy of the gas, and the rest of it gets converted to work.
Note:
Internal energy of a gas is defined as the sum of the kinetic energy and potential energy of the gas molecules. It also takes into account the vibrational, rotational and translational kinetic energy of the molecules.
Mathematically,
$\mathrm{\Delta Q}=\mathrm{\Delta}U+\mathrm{\Delta}W$
Q- Heat transfer between the system and the surroundings.
W- Work done by the system on the surroundings(or vice versa).
U- Change in internal energy of the gas
In differential form, it can be expressed as,
$dQ=dU+dW--\left(i\right)$
The first law of thermodynamics is a consequence of the law of conservation of energy. It talks about energy transfer from surroundings to the system (or vice versa).
The internal energy is a state function— it does not depend on the path taken by the system. Generally, the change in internal energy is given by,
$\mathrm{\Delta}U=\mu {C}_{v}\mathrm{\Delta T},\mathrm{\Delta T}-$change in temperature undergone by the gas.
Internal energy is also an extensive property–it depends upon the amount of matter present in the gas.
Cv- Specific heat capacity at constant volume.
Additionally, heat Q is a path function.
$\mathrm{\Delta Q}=\mu {C}_{P}\mathrm{\Delta T}$
Similarly, the work done by/ on the gas is also a path function.
Process |
Convention |
Heat added to the system |
Q>0 ie heat is positive |
Heat extracted from the system |
Q<0 ie heat is negative |
Work done by the system( volume increases) |
W>0 ie positive |
Work done on the system( volume decreases) |
W<0 ie negative |
When the temperature of the system rises |
U>0 ie positive |
When the temperature of the system falls |
U<0 ie negative |
The SI unit of heat, work and internal energy are Joule. Its dimensional formula is [ML_{2}T^{-2}].
Work in thermodynamics refers to work done by the system or on the system. Let us consider a gas being confined in a cylinder. The work done when the gas expands is given by,
dW=Fdx, where dx is the distance by which the gas pushes up the piston.
dW=P A dx, where A is the area of the cross section of the piston, P- pressure.
dW=P dV
The total work done can be found by integrating the above equation.
dW=ViVfP dV
Here Vf and Vi indicate the final and initial volumes of the system. During compression,
Vf<Vi hence work done is negative.
During expansion, work done is positive since Vf>Vi.
In other words, work done on the system is negative and work done by the system is positive.
Also, for constant pressure, work done, W= R TiTfdT= R(Tf-Ti).
Here is the number of moles of the gas and R is the gas constant.
Let us consider the following (pressure-volume) P-V diagram. The work done by the gas is given by the shaded region i.e area under the P-V diagram gives work done.
For an isothermal process, change in temperature is zero, ie. dT=0
dU= Cv dT=0. i.e there is no internal energy change in an isothermal process.
Hence, equation (i) becomes,
dQ=d W
I.e all the heat supplied to the gas gets converted into work.
For an adiabatic process, no heat is lost or gained in the process, hence dQ=0.
The system does not exchange heat with the surroundings.
$0=dU+dW\Rightarrow dU=-dW$
During an adiabatic compression, the work done dW is negative. Hence, dU>0.
During an adiabatic expansion, the work done dW is positive. Hence, dU<0.
During an isochoric process, there is no change in volume (dV=0), hence the work done, PdV is zero.
Hence from (i) we have dQ=dU.
I.e all of the heat supplied gets converted into internal energy.
For an isobaric process, the pressure is maintained constant.
1)For isobaric compression, we have W<0
Now U=Q-W
Now if heat is supplied, Q>0
$\Rightarrow \mathrm{\Delta U}>0$
2)For isobaric expansion, W>0
Now if heat is lost by the system, Q<0
$\mathrm{\Delta U}=\mathrm{\Delta Q}-\mathrm{\Delta W}$
$\Rightarrow \mathrm{\Delta U}<0$
Q. A gas is compressed by applying a constant pressure of 50 Nm-2 . Initially, its volume was 10 m3 but its volume becomes 4 m3 after compression. Energy of 100 J is supplied to the gas. Its internal energy is
(a)increased by 400 J (c) increased by 100 J
(b)increased by 200 J (d)decreased by 200 J
A. a
Given, heat supplied to the gas, $\mathrm{\Delta Q}=100J,{V}_{f}=4{m}^{3},{V}_{i}=10{m}^{3},P=50N{m}^{-2}$
Work done, $\mathrm{\Delta W}=P\mathrm{\Delta V}=P({V}_{f}-{V}_{i})=50(4-10)=-300J$
Negative sign indicates that work is done on the gas. When a gas is compressed, work is done on the system.
Internal energy, U=?
Applying the equation for first law of thermodynamics,
$\mathrm{\Delta Q}=\mathrm{\Delta U}+\mathrm{\Delta W}$
100= U-300
$\mathrm{\Delta U}=300+100=400J$
Hence, the internal energy of the gas increases by 400 J.
Q. An ideal monoatomic gas undergoes a process in which the heat content of the system remains unchanged. The change in temperature in the process is 300 K. Calculate the change in internal energy.
A.
Given that there is no change in heat contentQ=0, meaning it is an adiabatic process.
Change in temperature, T=300 K
0=U+W;
U=-W
- adiabatic constant for a monoatomic gas=53
$\mathrm{\Delta W}=\frac{R}{\gamma -1}\left(\mathrm{\Delta T}\right)=\frac{R}{\gamma -1}(300K)=\frac{R}{\frac{5}{3}-1}(300K)=\frac{3R}{2}(300)=450R=3739.5J$
Magnitude of change in internal energy,
$\mathrm{\Delta U}=3739.5J$
Q. The internal energy of a gas varies according to the relation, U=2P V. It undergoes expansion from V0 to 2 V0 against a constant pressure P0. Find the heat absorbed by the gas in the process.
(a)2 P0 V0 (b)4 P0 V0 (c)3 P0 V0 (d)8 P0V0
A. c
Internal energy change, $U=2PV\Rightarrow \mathrm{\Delta U}=2{P}_{0}\left(2{V}_{0}\right)-2{P}_{0}{V}_{0}=2{P}_{0}{V}_{0}$ is the change in internal energy,
Work done, $\mathrm{\Delta W}=P\mathrm{\Delta V}={P}_{0}(2{V}_{0}-{V}_{0})={P}_{0}{V}_{0}$
Heat absorbed, $\mathrm{\Delta Q}={P}_{0}{V}_{0}+2{P}_{0}{V}_{0}=3{P}_{0}{V}_{0}$
Q. A monoatomic gas is supplied a quantity of heat Q keeping the pressure constant. The work done by the gas will be
(a)$\frac{2Q}{3}$ (b)$\frac{3Q}{5}$ (c)$\frac{2Q}{5}$ (d)$\frac{Q}{5}$
A. c
Given, heat supplied is Q.
From the first law of thermodynamics, work done, W=Q- U
$\mathrm{\Delta Q}=\mu {C}_{p}dT=\mu dT\frac{5R}{2}=\frac{5\mu RdT}{2}$
$({C}_{p}=\frac{5R}{2}and{C}_{v}=\frac{3R}{2}$ for monoatomic gas)
$\mathrm{\Delta}U=\mu {C}_{v}dT=\frac{3\mu RdT}{2}$
$\mathrm{\Delta W}=\frac{5\mu RdT}{2}-\frac{3\mu RdT}{2}=\mu RdT=\frac{2}{5}\times \frac{5}{2}\mu RdT=\frac{2}{5}Q$
Q. What are the limitations of the first law of thermodynamics?
A. The first law states that the heat supplied to the gas is simply used to increase the internal energy of the gas and the rest is used to do work. But it does not talk about the direction of heat flow whatsoever. Additionally, it does not give information whether the process is spontaneous or not.
Q. Give an example of a machine that violates the first law of thermodynamics.
A. Perpetual motion machines violate the first law of thermodynamics. It will continue to work on and on without being disturbed–it does not need an external energy source for this to be possible. But in reality, a machine needs to be supplied with fuel, and it will continue to work only as long as the fuel is supplied.
Q. Does the human body obey the first law of thermodynamics?
A. Yes. The food we eat everyday is used for increasing the internal energy of the body. The rest of the energy is used in doing mechanical work. This is nothing but one more example of conservation of energy–the chemical energy of the food we consume being converted into other forms.
Q. Does the first law of thermodynamics apply to open systems?
A. Yes, the first law of thermodynamics is applicable to both open and closed systems. In open systems, both mass and energy are transferred from system to surroundings. So, when energy transfer occurs, the first law is applicable.