Question

Figure 18-56ashows a cylinder containing gas and closed by a movable piston.The cylinder is kept submerged in an ice–water mixture. The piston is quicklypushed down from position 1 to position 2 and then held at position 2 until the gas is again at the temperature of the ice–water mixture; it then is slowlyraised back to position 1. Figure 18-56bis a p-Vdiagram for the process. Ifof ice is melted during the cycle, how much work has been done onthe gas?

Short answer

Work done on the gas is $33.3\text{\hspace{0.17em}kJ}$.

Step-by-step solution

Stating the given data

1. Mass of ice is $m=100\text{\hspace{0.17em}gor}0.1\text{\hspace{0.17em}\hspace{0.17em}kg}$
2. Figure 18-56 is graph of pressure vs. volume.

Understanding the concept of work done

By expanding or contracting against a steady external pressure, gases can perform work. We use the concept of work done by the gas. We use the concept of work done. Using the equation of work done, we can find work done on the gas by the piston. We can find this using the amount of heat required to melt the given mass of ice.

Formula:

The work done on the gas due to thermodynamic process,$W=PdV$. (i)

The heat released by the gas due to fusion,$W=m{L}_f$. (ii)

Calculation of the work done on the gas

The work done on the gas by the piston is heat given to the ice-water mixture;so, when the piston is at position 2, the gas will have some heat due to the work done. This heat is transferred to ice to melt it;therefore, the heat given by the gas is equal to the heat absorbed by the ice.

We can write the work done on the gas using equation (ii) as

$\begin{array}{c}{W}_{\text{piston}}=(0.1\text{\hspace{0.17em}\hspace{0.17em}kg}\times 3.33\times {10}^5\text{\hspace{0.17em}J/kg})\text{(}\because {\text{specificheatoffusion,L}}_f\text{=}3.33\times {10}^5\text{J/kg)}\\ =3.33\times {10}^4\text{J}\end{array}$

Hence, the value of the work done on the gas is $33.3\text{\hspace{0.17em}kJ}$.