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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