A system undergoes a process which absorbed 0.5 kJ of heat and undergoing an expansion against an external pressure of 1 atm, during the process change in internal energy is 300 J. Then predict the change in volume (lit)
A.1
B.2
C.3
D.4

Hint: We know the formula for finding work done in case of expansion, is given by formula:
$W = - {P_{ext}} \times \Delta V$
In the above formula, if we get work done, then we can find change in volume easily as external pressure is given to us.
Now to get work done, we know the first law of thermodynamics, regarding law of conservation of energy. The law states that the total change in internal energy is the sum of net heat transfer and total work done. Mathematically we can write equation as:
\[\Delta U = Q + W\]

Complete step by step answer:
To begin answering, we have to write the formula of work done, as we know that it contains change in volume term.
$W = - {P_{ext}} \times \Delta V$ ----- equation 1
Here, W= work done by system, as it is expansion.
\[{P_{ext}} = 1atm\] (given)
\[\Delta V = \] change in volume (this is what we have to find out)
Substitute values in equation 1, we get
\[W = - (1) \times \Delta V{\text{ atm}}{\text{.L}}\]
Now, since other parameters are given in joules, we should convert this work done in joules.
1 atm.L = 101.3 Joules
\[\therefore W = - 101.3 \times \Delta V\] ---- equation 2
Now, the mathematical form of first law of thermodynamics is:
\[\Delta U = Q + W\] ----- equation 3
Here, \[\Delta U = 300J\] = internal energy change (Given in question)
Q=0.5 kJ = 500 J =heat supplied to system (Given in question)
On substituting equation 2, values of internal energy change and heat supplied, in equation 3, we get:
\[300 = 500 - 101.3 \times \Delta V\]
Taking, 500 on left side, we get
\[300 - 500 = - 101.3 \times \Delta V\]
On simplifying and cancelling negative sign from both sides we get:
\[200 = 101.3 \times \Delta V\]
Taking numbers on one side, 101.3 will get divided and we get:
\[\dfrac{{200}}{{101.3}} = \Delta V\]
Thus we get value of change in volume as:
\[\therefore \Delta V = 1.97 \approx 2L\]
We have to take the nearest whole number as options are given in whole numbers.
Thus the change in volume is 2 litres.
Hence, the correct option is (B).


Note:
Take care of negative and positive signs in case of Work done for expansion and compression, as we know work done for expansion is negative, as volume increases in expansion and vice versa for compression. Also one should be careful for conversion of work done from atm.L to Joules.