Answer to Question #87602 in Molecular Physics | Thermodynamics for Lesly

Question

2.9 cm3 of water is boiled at atmospheric pressure to become 5271.2 cm3 of steam, also at atmospheric pressure.
1. Calculate the work done by the gas during this process. The latent heat of vaporization of water is 2.26 × 106 J/kg. Answer in units of J.
2. Find the amount of heat added to the water to accomplish this process. Answer in units of J.
3. Find the change in internal energy. Answer in units of J.

Answer 1

$$W =\int pdV$$

As pressure is consatnt:

$$W = p\times \Delta V = 101325 Pa \times (5271.2\times 10^{-6} m^3 - 2.9\times10^{-6} m^3) =533.81 J$$

2. Heat absorbed by the water during vaporization process:

$$Q = m\times \Delta H_{vap}$$

$$m(water) = d(water)\times V(water) = 958.4 \frac{kg}{m^3}\times 2.9\times 10^{-6} m^3 =2.779 \times 10^{-3}kg$$

$$Q = 2.779 g \times 10^{-3} kg \times 2.26\times 10^6\frac{J}{kg} =6281.35 J$$

3. Chage in internal energy of a system is equal to heat added to the system (= heat absorbed) minus the work done by the system (by gas):

$$\Delta U = Q-W = 6281.35 J - 533.81 J =5747.54 J$$