- Transfer of heat and performance of work are two mean of adding or subtracting energy from a system.

- On transfer of energy, system is said to have undergone a change in internal energy.

- Thus the sum of heat put into the system plus work done on the system equals increase in internal energy of the system for any process.

if, $U_1$ is internal energy of state 1 and $U_2$ is internal energy of state 2 than change in internal energy would be

$\Delta U=U_2 - U_1$

- If W is the work done by the system on its surroundings then -W would be the work done on the system by the surroundings .

- If Q is the heat put into the system then,

Applying general law of conservation of energy $Q+(-W)= \Delta U$

usually written as

$Q=\Delta U + W$ --(1)

- Equation (1) is then know as first law of thermodynamics and it can be applied when Q, W and U are expressed in same units.

- Sometimes, we denote Heat and work by $\Delta Q$ and $\Delta W$,then
$\Delta Q=\Delta U + \Delta W$ --(2)

- $\Delta Q$ or Q is positive when heat is given to the system and Q is negative when heat is taken from the system

- $\Delta W$ or W is positive when system expands and does work on surroundings

- Hence we may say that when a certain amount of heat Q is given to the system then some part of it is used in increasing internal energy $\Delta U$ of the system while remaining part leaves the system in form of work done by the system on its surroundings.

- From equation 4 we see that first law of thermodynamics is a statement of conservation of energy stated as

' The energy put into the system equals the sum of the work done by the system and the change in internal energy of the system'

- If the system undergoes any process in which $\Delta U=0$ i.e., charge in internal energy is zero then from (1)

$Q = W$

or

$\Delta Q = \Delta W$

that is heat supplied to the system is used up entirely in doing work on the surroundings.

- If system reach from State A to state B by executing number of steps and $Q_1$, $Q_2$, $Q_3$ and $W_1$, $W_2$, $W_3$ are heat and work in each step, we can write the change of internal energy between state A and B as

$ \Delta U= Q_1 + Q_2 + Q_3 - (W_1 + W_2 + W_3$

- If the system returns to its original state after executing number of steps, the $ \Delta U=0$ and

$Q_1 + Q_2 + Q_3 = W_1 + W_2 + W_3$

- Equation (1) can be written as

$ \Delta U= Q -W $

Now since U being a thermodynamic state variable,it value does not depend on the path taken. but Q and W depends on the path . From the equation we can say that Combination of Q - W would be independent of the path

A gas is given 50 Calorie of heat and Gas does the 20 J of work in the expansion resulting from heat given. Find the increase in the Kinetic energy of the gas in the process?

Q= 50 Calorie= 209 J

W= 20 J

From first law of Thermodynamics

$ \Delta U= Q -W $

$= 209 -20 = 189 J$

- Introduction
- Concept of Heat
- P-V Indicator Digram
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- Work done by Gas in volume changes
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- Internal Energy
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- First Law of Thermodynamics
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- Specific Heat Capacity of Ideal GAS
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- Thermodynamic Processes
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- Quasi static Processes
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- Isothermal Process
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- Adiabatic Process
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- Isochoric process
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- Isobaric process
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- Cyclic process
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- Work done in Isothermal process
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- Work done in an Adiabatic process
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- Heat Engine and efficiency
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- Principle of a Refrigerator
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- Second law of thermodynamics
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- Reversibility and irreversibility
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- Carnot's Heat Engine
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- Carnot Theorem
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- Solved Examples

- Thermodynamics Questions
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- Multiple Choice Questions
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- P-V diagram Problems and Solutions
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- Carnot Cycle Problems

Class 11 Maths Class 11 Physics Class 11 Chemistry

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