Question

a) 1kg of water at 0oC(degree Celsius) is fully converted into steam at 100°C at normal pressure. Calculate the change in entropy. The specific heat capacity of water is
4.18X10^3 JKg^-1K^-1 and latent heat of vaporisation is 2.24x10^6 JKg^-1.
b) Calculate the work done by one mole of a van der Waals’ gas if during its isothermal
expansion its volume increases from 1m^3 to 2m^3 at a temperature 300K.
Take a = 1.39X10^-6 atm m^6 mol^-2 and b = 39.1X10^-6 m^3 mol^-1.

Answer 1

Answer on Question#39780, Physics, Molecular Physics

a) $1\mathrm{kg}$ of water at 0oC (degree Celsius) is fully converted into steam at $100{}^{\circ}\mathrm{C}$ at normal pressure. Calculate the change in entropy. The specific heat capacity of water is $4.18 \times 10^{-3} \mathrm{JKg}^{-1} \cdot 10^{-1}$ and latent heat of vaporization is $2.24 \times 10^{-6} \mathrm{JKg}^{-1}$ .

Solution

Entropy is the reversible enthalpy change of a process divided by $T$ . Assuming the heat changes in the given situations to be reversible in nature:

$$\Delta S = \Delta S _ {h e a t i n g} + \Delta S _ {v a p o r i z i n g} = \int \left(\frac {d q _ {r e v}}{T}\right) + \frac {q _ {2}}{T _ {s t e a m}}.$$

For heating up of water,

$$q = m \cdot c \cdot d T,$$

where $c$ is the specific heat of water.

Thus,

$$\Delta S _ {h e a t i n g} = \int m c \left(\frac {d T}{T}\right) = m c \ln \frac {T _ {2}}{T _ {1}}.$$ $$q _ {2} = m \cdot L,$$

with $L$ the latent heat of vaporization.

Therefore,

$$\Delta S = m c \ln \frac {T _ {2}}{T _ {1}} + \frac {m \cdot L}{T _ {2}} = 1 \mathrm {k g} \cdot 4. 1 8 \cdot 1 0 ^ {3} \frac {\mathrm {J}}{\mathrm {k g} \mathrm {K}} \cdot \ln \left(\frac {3 7 3}{2 7 3}\right) + \frac {1 \mathrm {k g} \cdot 2 . 2 4 \cdot 1 0 ^ {6}}{3 7 3 \mathrm {K}} \frac {\mathrm {J}}{\mathrm {k g}} = 7. 3 1 \frac {\mathrm {k J}}{\mathrm {K}}.$$

Answer: 7.31 kJ

b) Calculate the work done by one mole of a van der Waals' gas if during its isothermal expansion its volume increases from $1\mathrm{m}^{\wedge}3$ to $2\mathrm{m}^{\wedge}3$ at a temperature 300K. Take $a = 1.39 \times 10^{\wedge} - 6$ atm $\mathrm{m}^{\wedge}6 \mathrm{~mol}^{\wedge} - 2$ and $b = 39.1 \times 10^{\wedge} - 6 \mathrm{~m}^{\wedge}3 \mathrm{~mol}^{\wedge} - 1$ .

Solution

The work done by one mole of gas is

$$W = \int_ {v _ {1}} ^ {v _ {2}} p d v,$$

where $v$ is the molar volume.

The Van der Waals equation of state for 1 mole of gas is:

$$\left(p + \frac {a}{v ^ {2}}\right) \cdot (v - b) = R T.$$

From this equation we can obtain that:

$$p = \frac{RT}{v - b} - \frac{a}{v^2}.$$

The work done by one mole of a van der Waals' gas is

$$W = \int_{v_1}^{v_2} \left( \frac{RT}{v - b} - \frac{a}{v^2} \right) dv = \int_{v_1}^{v_2} \left( \frac{RT}{v - b} \right) dv - \int_{v_1}^{v_2} \left( \frac{a}{v^2} \right) dv = RT \ln \frac{v_2 - b}{v_1 - b} + a \left( \frac{1}{v_2} - \frac{1}{v_1} \right).$$

We can obtain $a$ in SI units:

$$a = 1.39 \cdot 10^{-6} \cdot 1.01325 \cdot 10^5 \cdot \mathrm{N} \cdot \mathrm{m}^{-2} \cdot \mathrm{m}^6 \cdot \mathrm{mol}^{-2} = 0.141 \, \mathrm{N} \cdot \mathrm{m}^4 \cdot \mathrm{mol}^{-2}.$$

So

$$W = 8.31 \cdot 300 \cdot \ln \frac{2 - 39.1 \cdot 10^{-6}}{1 - 39.1 \cdot 10^{-6}} + 0.141 \left( \frac{1}{2} - \frac{1}{1} \right) = 1.73 \, \mathrm{kJ}.$$

Answer: $1.73 \, \mathrm{kJ}$.