# First law of thermodynamics

## First law of thermodynamics

• 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)

### Some Important things to remember about First law of Thermodynamics

• $\Delta Q$ 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

## Solved Example

Question 1
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?
Solution
Q= 50 Calorie= 209 J
W= 20 J
From first law of Thermodynamics
$\Delta U= Q -W$
$= 209 -20 = 189 J$

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