Question

When 1 g of water at \[{0^o}\;C\] and \[1\; \times \;{10^5}\;N/{m^2}\;\] pressure is converted into ice of volume \[1.091\;c{m^3}\] , the external work done will be:
A) 0.0091 joule
B) 0.0182 joule
C) −0.0091 joule
D) −0.0182joule

Answer 1

Hint:
No change in temperature hence it is an isothermal process. An isothermal process is a process where the temperature of the system remains constant. That means that no heat is exchanged with the surroundings. The total amount of energy in a system is constant. Energy can be transferred between systems, but the total amount of energy remains the same.


Complete step by step solution:

By looking at the question carefully we see that water is at \[{0^o}\;C\] and also the ice is at \[{0^o}\;C\] temperature. Hence, no change in temperature. So, the process is isothermal because only change in states happens. Let’s discuss work done in an isothermal process.
An isothermal process is a process where the temperature of the system remains constant. That means that no heat is exchanged with the surroundings.
Now let’s solve the question.
According to the first law of thermodynamics, the change in internal energy is the function only of temperature.
Hence, \[\vartriangle U = f\left( T \right)\] where T is the temperature, \[\vartriangle U\] is change in internal energy.
If temperature is constant then \[\vartriangle U = 0\].
Now, \[\vartriangle Q = \vartriangle U + W\] here \[\vartriangle Q\]is the net heat transfer and W is network done.
\[ \Rightarrow \vartriangle Q = W\]
For an isothermal process, \[W = \smallint PdV\] here P is pressure and V is volume.
We know that the density of water \[1gm/c{m^3}\]. Hence the volume of 1 gm water \[1c{m^3}\]. Hence,
\[W = p\vartriangle V\]$ = 1 \times {10^5} \times (1.091 - 1) \times {10^{ - 6}}{\text{ Joule}}$
$ = 0.091 \times {10^{ - 1}} = 0.0091{\text{joule}}$
Hence the correct option is A.



Therefore, option (A) is the correct option.



Note:
When a gas is compressed isothermally then heat is rejected from the system. The first law of thermodynamics, the change in internal energy is the function only of temperature. The mathematical expression of laws of thermodynamics is , \[\vartriangle Q = \vartriangle U + W\] here \[\vartriangle Q\]is the net heat transfer and W is net work done , \[\vartriangle U\]= change in internal energy. Work done on an isothermal process is \[W = \smallint PdV\].