Question

Given, \[1g\] of water (volume \[1c{{m}^{3}}\]) becomes \[1671c{{m}^{3}}\] of steam when boiled at a pressure of \[1atm\]. The latent heat of vaporization is \[540cal/g\], then the external work done is :
( \[1atm=1.013\times {{10}^{5}}N/{{m}^{2}}\])
A. \[499.9\text{ }J\]
B. \[40.3\text{ }J\]
C. \[169.2\text{ }J\]
D. \[128.57\text{ }J\]

Answer 1

Hint: We can solve this using the first law of thermodynamics. It states that the energy given to a system will be used to change the internal energy and work done. This work is found by change in volume.

Formula used: \[\Delta U=\Delta Q-\Delta W\]
                                   \[\Delta W=P\Delta V\]

Complete answer: In a thermodynamic system with volume change, The amount of external work done is given by the product of pressure and change in volume.
Now, we have the formula for work done as,
             \[\Delta W=P\Delta V\]
Here,\[\Delta W\]is the work done,\[P\] is pressure and \[\Delta V\] is change in volume
In the question the volume of water and steam is given. So the change in volume \[\Delta V\] can be found as,
            \[\Delta V={{V}_{Steam}}-{{V}_{Water}}\]
             \[\begin{align}
  & {{V}_{steam}}=1671c{{m}^{3}} \\
 & {{V}_{water}}=1c{{m}^{3}} \\
\end{align}\]
\[\Rightarrow \Delta V=(1671-1)c{{m}^{3}}=1670c{{m}^{3}}=1670\times {{10}^{-6}}{{m}^{3}}\] (here, \[c{{m}^{3}}\]is converted to\[{{m}^{3}}\])
Now, to find\[\Delta W\], pressure is given as\[1atm=1.013\times {{10}^{5}}N/{{m}^{2}}\]
      \[\begin{align}
  & \Delta W=P\Delta V \\
 & \Rightarrow \Delta W=1.013\times {{10}^{5}}\Delta V \\
\end{align}\]
              \[\begin{align}
  & \Delta W=1.013\times {{10}^{5}}\times 1670\times {{10}^{-6}} \\
 & \Delta W=1.013\times 167 \\
 & \Delta W=169.17J\approx 169.2J \\
\end{align}\]
The latent heat doesn’t take part in the external work done.it only take part in increase in internal energy which can be shown by,
               \[\Delta U=\Delta Q-\Delta W\]
Here,\[\Delta U\] is a change in internal energy and \[\Delta Q\] is given energy which could be found as \[mL\] where \[L\] is latent heat.
\[\Rightarrow \Delta U=mL-\Delta W\] (\[1\text{ }kcal\text{ }=\text{ }4180j\])
      \[\begin{align}
  & \Delta U=1\times 540-(169.2\div 4.18) \\
 & \Delta U=540-40.47 \\
 & \Delta U\approx 500cal \\
\end{align}\]
Increase in internal energy is \[500cal\].
So the correct answer is option c

Note: This question may seem a little bit confusing as they have given latent heat in question but asked only about external work done. So we must be aware that the constituents in the first law of thermodynamics must be able to distinguish between them.