PV diagram is a graph between pressure P and volume V of the system.

First law of thermodynamics states that change in internal energy of the system is equal to heat supplied to the system (Q) minus the work done by the system (W) on the surrounding.
ΔU = Q-W

One calorie is defined as the amount of energy required to raise 1 gram of an object by 1^oc.

Wien's law states that, the wavelength of maximum intensity of emission of a black body radiation is inversely proportional to the absolute temperature of the black body.
\({ \lambda }_{ m } \)∝\(\frac { 1 }{ T } \) 
\({ \lambda }_{ m }=\frac { b }{ T } \)

 One mole of any substance is the amount of that substance which contains Avogadro number (N_A) of particles such as atoms or molecules.

COP = \(\beta\) = \(\frac{Q_L}{W}\)
W = \(\frac{QC}{COP}=\frac{200}{3}\) =  66.67J

The steam engine is not a Carnot engine, because all the process involved in the steam engine are not perfectly reversible. But we can calculate the maximum possible efficiency of the steam engine by considering it as a Carnot engine.
\(\eta =1-\frac { { T }_{ L } }{ { T }_{ H } } =1-\frac { 300K }{ 523K } =0.43\)
The steam engine can have maximum possible 43% of efficiency, implying this steam engine can convert 43% of input heat into useful work and remaining 57% is ejected as heat. In practice the efficiency is even less than 43%.

When the water is heated from 30°C to 60°C, there is only a slight change in its volume. So we can treat this process as isochoric. In an isochoric process the work done by the system is zero. The given heat supplied is used to increase only the internal energy.
ΔU = Q = ms_v ΔT
The mass of water = 500 g = 0.5 kg
The change in temperature = 30K
The heat Q = 0.5\(\times\)4184\(\times\)30 = 62.76 kJ

Consider a container of gas with volume V, pressure P and temperature T. If we add sand particles one by one slowly on the top of the piston, the piston will move inward very slowly. This can be taken as almost a quasi-static process. It is shown in the figure

Sand particles added slowly- quasi-static process

We can take air molecules in the tire as an ideal gas. The number of molecules and the volume of tire remain constant. So the air molecules at 27°C satisfies the ideal gas equation P₁V₁ = NkT₁ and at 39°C it satisfies P₂V₂ = NkT₂
But we know
V₁ = V₂ = V
\(\frac { { P }_{ 1 }V }{ { P }_{ 2 }V } =\frac { Nk{ T }_{ 1 } }{ Nk{ T }_{ 2 } } \)
\(\frac { { P }_{ 1 } }{ { P }_{ 2 } } =\frac { { T }_{ 1 } }{ { T }_{ 2 } } \)
\( { P }_{ 2 } =\frac { { T }_{ 1 } }{ { T }_{ 2 } } { P }_{ 1 }\)
\({ P }_{ 2 }=\frac { 312K }{ 300K }\times 240\times{ \ 10 }^{ 3 }Pa=249.6Pa\qquad \)

\(\mathrm{COP}=\frac{T_{L}}{T_{H}-T_{L}}\)
\(\mathrm{T}_{\mathrm{L}}=-12+273 =261 \mathrm{~K} \)
\(5=\frac{261}{T_{H}-261} \)
\(\therefore 5\left(T_{H}-261\right) =261 \)
\(5 \mathrm{~T}_{\mathrm{H}}-1305 =261 \)
\(5 \mathrm{~T}_{\mathrm{H}} =261+1305 =1566 \)
\(\therefore T_{H} =\frac{1566}{5} \)
\(=313.2 \mathrm{~K} \)
\(\mathrm{~T}_{\mathrm{H}}=313.2-273 =40.2^{\circ} \mathrm{C} \)

\(\mathrm{T}_{2} =0^{\circ} \mathrm{C}+273=273 \mathrm{~K} \)
\(\mathrm{~T}_{1}=100^{\circ} \mathrm{C}=100+273=373 \mathrm{~K} \)
\(\eta =1-\frac{T_{2}}{T_{1}} \)
\(=1-\frac{273}{373} \)
\(=\frac{373-273}{373} \)
\(=\frac{100}{373}=0.26809 \times 100 \)
\(=26.809 \% \)

\(\Delta Q=\Delta U+\Delta W\)
When, exercise is performed internal energy is utilised (\(\Delta\)U < 0) so it will decrease. But as intemal energy is utilised, exercise ( \(\Delta\)W ) will be more efficient.
\(\therefore \Delta W>0 \ \therefore \Delta U <0 \mathrm{~W}>0\)

Process 1 to 2 = increase in volume. So heat must be added.
Process 2 to 3 = Volume remains constant. Increase in temperature. The given heat is used to increase the internal energy.
Process 3 to 1 = Pressure remains constant. Volume and Temperature are reduced. Heat flows out of the system. It is an isobaric compression where the work is done on the system.
In the process marked as 1-2, the gas undergoes isothermal 'expansion. It receiver certain amount of heat from the outside of spends all this heat in doing work the internal energy of the gas remains unchanged.
In the process marked as 2-3, the gas is heated isochorically (at constant volume). Since its volume does not change, no work is done. The internal energy of the gas is increased only due to the heat transferred to the gas from the outside.
In the process exhibited by 3-1, the gas is compressed isobarically (at constant pressure) of its temperature drops work is done on the gas, but its internal energy is reduced. This means the gas intensively gives up heat to the medium.

Heat engine is a device which takes heat as input and converts this heat in to work by undergoing a cyclic process.

A black body is the one which absorbs completely heat radiations of all wavelengths when heated. A black body neither reflects nor transmits any radiation. Its absorptive power is unity.

Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature.

Newton's law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac{d Q}{d t} \propto-\left(\mathrm{T}-\mathrm{T}_{s}\right)\)   ...(1)
The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,
T = Temperature of the object
T_s = Temperature of the surrounding

From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.
Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount dT in time dt, then the amount of heat lost is,
dQ = msdT ........(2)
(iv) Dividing both sides of equation (2) by dt
\(\frac { dQ }{ dt } =\frac { msdT }{ dt } \)..........(3)
From Newton's law of cooling
\(\frac { dQ }{ dt } \propto -(T-{ T }_{ s })\)
\(\frac { dQ }{ dt } =-\alpha (T-{ T }_{ s })\)   ...(4)
Where a is some positive constant.
From equation (2) and (4)
-\(\alpha\) (T - T_s) = \(md\frac { dt }{ dt } \)
\(\frac { dt }{ T-{ T }_{ s } } =\frac { a }{ ms } dt\)   ....(5)
Integrating equation (5) on both sides,
\(\int _{ 0 }^{ \infty }{ \frac { dt }{ T-{ T }_{ a } } =-\int _{ 0 }^{ 1 }{ \frac { a }{ ms } dt } } \)
ln (T - T_s) = \(-\frac { a }{ ms } t+{ b }_{ 1 }\)
Where b₁ is the constant of integration taking exponential both sides, we get
\(\mathrm{T} =T_{s}+b_{2} e^{-\frac{a}{m s} t} \)
here \(\mathrm{~b}_{2} =e^{b_{1}}=\text { constant }\)

When the gas is kept at constant temperature, the pressure of the gas is inversely proportional to the volume. \(P \propto \frac{1}{V}\). It was discovered by Robert Boyle (1627-1691) and is known as Boyle's law.
When the gas is kept at constant pressure, the volume of the gas is directly proportional to absolute temperature. \(V \propto T\). It was discovered by Jacques Charles (1743-1823) and is known as Charles' law.
By combining these two equations we have
PV = CT. Here C is a positive constant.
We can infer that. C is proportional to the number of particles in the gas container by considering the following argument. If we take two containers of same type of gas with same volume V, same pressure P and same temperature T, then the gas in each container obeys the above equation. PV = CT. If the two containers of gas is considered as a single system, then the pressure and temperature of this combined system will be same but volume will be twice and number of particles will also be double as shown in Figure.

For this combined system, V becomes 2V, so C should also double to match with the ideal gas equation \(\frac { P(2V) }{ T } =2C\). It implies that C must depend on the number of particles in the gas and also should have the dimension of  \(\left[ \frac { PV }{ T } \right] ={ JK }^{ -1 }\). So we can write the constant C as k times the number of particles N.
Here k is the Boltzmann constant (1.381\(\times\)10^-23 JK^-1) and it is found to be a universal constant.
So the ideal gas law can be stated as follows
PV = NkT  ....(1)
The equation (1 )can also be expressed in terms of mole.
Suppose if a gas contains \(\mu\) mole of particles then the total number of particles can be written as
\(\mathrm{N}=\mu \mathrm{N}_{\mathrm{A}}\)    .....(2)
where N_A is Avogádro number \(\left(6.023 \times 10^{23} \mathrm{~mol}^{-1}\right)\)
Substituting for N from equation (2), the equation (1) becomes
\(\mathrm{PV}=\mu N_{A} k T\)
Here \(\mathrm{N}_{\mathrm{A}} \mathrm{k}=\mathrm{R}\) called universal gas constant and its value is \(8.314 \mathrm{~J} / \mathrm{mol}. \mathrm{K}\).
So the ideal gas law can be written for $\mu$ mole of gas as
\(\mathrm{PV}=\mu R T\)
This is called the equation of state for an ideal gas. It relates the pressure, volume and temperature of thermodynamic system at equilibrium.