Question

Two kg of water is converted into steam by boiling at atmospheric pressure. The volume change from $2 \times {10^{ - 3}}{m^3}$ to $3.34{m^3}$ . The work done by the system is about
A. $ - 340KJ$
B. $ - 170KJ$
C. $170KJ$
D. $340KJ$

Answer 1

Hint: When water is converted to steam heat energy is required to break the intermolecular bonds in the liquid state and thereby volume of steam increases as compared to liquid water. So, here we can use the pressure-volume work equation to get the actual amount of work done for this process.

Formula used:
The formula used in this problem is: -
$Workdone = W = P\Delta V$

Complete answer:
Since, the water is converted into steam at atmospheric pressure according to given question and we know that the value of atmospheric pressure is ${P_a} = 1.013 \times {10^5}Pa$
During the boiling of water, the volume changes as ${V_1} = 2 \times {10^{ - 3}}{m^3}$ and ${V_2} = 3.34{m^3}$(given)

Therefore, the change in volume can be given as $\Delta V = \;\;{V_2} - {V_1} = 3.34 - 2 \times {10^{ - 3}} = 3.338{m^3}$

Now, we know that Pressure-Volume Work in thermodynamics is defined as: -
$Workdone = W = P\Delta V$

Substituting the value of $P$ and $\Delta V$ in the above expression, we get
$ \Rightarrow W = 1.013 \times {10^5}\left( {3.338} \right)$
$ \Rightarrow W = 3.381 \times {10^5} = 338.1 \times {10^3}J$

Approximating the values, we get
$ \Rightarrow W = 338.1KJ \approx 340KJ$ $\left( {\therefore 1KJ = {{10}^3}J} \right)$
Thus, Approximately, the work done by the system is about $340KJ$.

Hence, the correct option is (D) $340KJ$ .

Note: In a isochoric process is actually a thermodynamic process under this condition work done will be zero because in that process volume of gas remains same and if we putting this value into the pressure-volume work equation we will get the value of work done as zero.