According to the first law of thermodynamics, when some amount of heat is given to the system capable of doing external work, then the quantity of heat it absorbs is equal to the sum of the increase in the internal energy of the system due to rise in temperature and the external work done by the system during expansion.

We represent the first law of thermodynamics by the equation:

ΔU = ΔQ - ΔW

This is the first law of thermodynamics formula.

Where

ΔU = Change in internal energy of the thermodynamic system

ΔQ = Heat given to the system

ΔW = Work done on the system

In differential form, the above equation is:

dU = dQ - dW

The first law of thermodynamics is based on the principle of the law of conservation of energy. This means energy can neither be created nor be destroyed but can transform into various forms with no loss of energy.

When a system changes from a given initial state to a given final state, both dQ and dW depend on the nature of the process. However, the quantity dQ - dW = dU is the same for all the processes.

Let’s study the formulas corresponding to each term, i.e., dU, dQ, and dW.

A. dU

At the molecular level, this energy is equal to the sum of kinetic energies and potential energies because of three types of motion of molecules, that is:

U = KETranslational motion + KEVibrational motion + KERotational motion + PEForce of attraction of electron and nucleus +Binding energy of molecules

Since it is difficult to calculate U at a microscopic level, so we study the change in internal energy dU.

dU is a state function, i.e., it depends on the initial and final state of the system.

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1. dU = nCVΔT

(dU is independent of the process, i.e. this formula remains the same every time)

Where,

n = number of moles

ΔT = Small temperature at which the gas is heated

Here, CV = molar specific heat capacity at constant volume

B. dQ

Heat added to the system is +ve, and

The heat released by the system = - ve.

At constant pressure, dQ = nCPΔT

(CP is the molar specific heat capacity of the gas at constant pressure)

At constant volume, dQ = nCVΔT

C. dW

Work is done on the system = +ve

Work is done by the system = - ve

W = - PExternalΔT

1. Isothermal process

In an isothermal process, the temperature of an ideal gas remains constant.

So, dQ = dU + dW ⇒ dQ = dW

This means heat supplied to the system is used to do work against the surroundings.

2. Melting process

When a solid melts to liquid, its internal energy increases.

Let m = mass of liquid

L = latent heat of the solid

Amount of heat absorbed by the system, dQ = mL

A small amount of expansion occurs, i.e., ΔV = 0 ⇒ dW = PΔV = 0

So, dQ = dU + dW ⇒ dU = mL

Thus, internal energy increases during the melting process.

3. Mayer’s formula or the relationship between CP and CV

Consider one mole of gas enclosed in a thermodynamic system fitted with a frictionless piston.

Let P, V, T be the pressure, volume, and temperature of the gas

We know that the formula for one mole of gas heated at a low temperature and constant volume is CVdT.

At constant volume, dW = PdV = 0.

dQ = dU + dW ⇒ dU = CVdT

If the gas is heated at constant pressure, then, dQ’ = CPdT

The work done, dW’ = PdV

Now, putting these values in the equation dQ’ = dU’ + dW’

⇒ CPdT = dU’ + PdV

Here, dU = dU’ = CVdT (= dT is same)

⇒ CPdT = CVdT + PdV

= (CP - CV)dT = PdV ….(1)

Now, using the ideal gas law equation, i.e., PV = RT and differentiating both the sides, we get

PdV = RdT (P, R = constant)...putting in (1)

(CP - CV)dT = RdT

⇒ CP - CV = R

1. The first law of thermodynamics does not show the direction in which the change occurs.

During winters, when we apply brakes to stop our car, work done against friction converts into heat. However, when it cools down, it does not start moving with the conversion of all its heat energy. This means that the heat stored should have been utilized in starting the car. It didn’t happen, so where is the heat gone? This becomes a limitation of this law.

2. Water flows down the Himalayas, where the potential energy converts to the kinetic energy, but we have not seen the reverse of this natural process.

3. Consider heating a room by the flow of electric current through an electric resistor. Transferring heat from the room will not cause electrical energy to be generated through the wire.

So, our observations and experiences tell us that, there appears to be no restriction on the conversion of mechanical work into heat. The heat is not converted by itself, an external agency i.e., a heat engine is required for this purpose.

FAQ (Frequently Asked Questions)

Q1: 750 calories of heat is added to a system and the system performs 550 calories of work. Calculate the change in Internal Energy.

Ans: Here, ΔQ = 750 Cal, ΔW = 550 Calories

We know that ΔU = ΔQ - ΔW

So, ΔU = (+) 750 - 550 = + 200 Calories is the increase in internal energy.

Q2: What is the C_{P}/C_{V} Ratio?

Ans: The C_{P}/C_{V }ratio is the ratio of specific heats, known as the adiabatic index.

Q3: 0.8 moles of O_{2} are heated from 30 to 150°C at constant pressure.

Calculate:

(i) Q

(ii) ΔU

(iii) W

(Given C_{P} = 7 cal/mol-K)

Ans: Here, n = 0.8, ΔT = 150 - 30 = 120 °C

(i) Q

We know at constant pressure,

Q = nC_{P}ΔT = 0.8 x 7 x 120 = 672 Cal

(ii) ΔU

dU = nC_{V}ΔT

Since CV = C_{P}_{ }- R

For a diatomic molecule, R = 2 cal/mol-K

⇒ C_{V}_{ }= 5 cal/mol-K ⇒ dU = 0.8 x 5 x 120 = 480 Cal

(iii) W

dW = dU - dQ ⇒ 480 - 672 = - 592 Cal, i.e. work is done by the system.

Q4: A sample of an ideal gas (γ = 1.4) is heated at constant pressure. If 120 J is supplied to it, then find dU and dW.

Ans: Here, dQ = 120 J, n =1

dU = C_{V}dT = C_{P}(C_{v}/C_{p})dT = C_{pdT}/γ = dQ/γ= 86 J

Now, dW = dQ - dU = 120 - 86 = + 34 J, i.e. work is done on the system.