Question

Question 01:
2.0g of Nitrogen gas is at 270C in a fixed volume. If 20% of its total internal energy is due to rotation what is the average velocity of nitrogen molecules?
What will be the temperature 2.0 g of Helium gas in a fixed volume if the average velocity of its molecules is 20 m/s ?
Plot schematic of Cv v/s T curve based for nitrogen and helium gas based on Kinetic theory.

Question 02:
200g of melting ice is introduced to a huge lake at 270C. Find the total entropy change of ice-lake system.
200g of melting ice is introduced to a huge lake at 0.20C. Find the total entropy change of ice-lake system.
Can you draw any conclusion based on result? (Hint: tends to reversible, irreversible etc)

Answer 1

Answer on Question #65269 -Physics - Molecular Physics - Thermodynamics

Condition:

**Question 01:**

2.0g of Nitrogen gas is at 270C in a fixed volume. If 20% of its total internal energy is due to rotation what is the average velocity of nitrogen molecules?

What will be the temperature 2.0 g of Helium gas in a fixed volume if the average velocity of its molecules is 20 m/s ?

Plot schematic of Cv v/s T curve based for nitrogen and helium gas based on Kinetic theory.

**Question 02:**

200g of melting ice is introduced to a huge lake at 270C. Find the total entropy change of ice-lake system.

200g of melting ice is introduced to a huge lake at 0.20C. Find the total entropy change of ice-lake system.

Can you draw any conclusion based on result? (Hint: tends to reversible, irreversible etc)

Solution:

**Question 01:**

Nitrogen gas is diatomic gas so $N_{2}$ has 5 degrees of freedom so total internal energy is $U = \frac{5}{2} kT * N$, where $k$ is Boltzmann constant ($k = 1.38 \times 10^{-23} \, \text{J/K}$), $T = 270C = 543 \, \text{K}$, $N$ is quantity of molecules.

$N = \frac{m}{M} N_{A}$, where $m = 2.0 \, \text{g}$, $M$ is molar mass (Molecular Weight):

$M(N_{2}) = 28 \frac{g}{mol}$, $N_{A}$ is Avogadro constant ($N_{A} = 6,023 * 10^{23} \, \text{mol}^{-1}$).

$$N = \frac{2}{28} * 6,023 * 10^{23} = 0,43 * 10^{23} = 4,3 * 10^{22}$$ $$U = \frac{5}{2} * 1,38 * 10^{-23} * 543 * 4,3 * 10^{22} = 805,54 \, \text{J}$$

If 20% of its total internal energy is due to rotation then 80% is the average energy of motion:

$$E = 0,8 * 805,54 = 644,4 \, \text{J}$$

On the other hand: $N * E_{0}$, where $N$ is quantity of molecules and $E_{0}$ is kinetic energy of one molecule.

$$E_{0} = \frac{E}{N} = \frac{644,4}{4,3 * 10^{22}} = 149,86 * 10^{-22} = 1,5 * 10^{-20} \, \text{J}$$

On the other hand: $E_0 = m_0 \frac{V^2}{2}$, where $V$ is the average velocity, $m_0$ - molecular mass:

$$m_{0} = 28 * 1,66 * 10^{-27} = 4,65 * 10^{-26} \, \text{kg}$$ $$V = (\frac {2 E _ {0}}{m _ {0}}) ^ {1 / 2} = \left(\frac {2 * 1 , 5 * 1 0 ^ {- 2 0}}{4 , 6 5 * 1 0 ^ {- 2 6}}\right) ^ {1 / 2} = (0, 6 4 5 * 1 0 ^ {6}) ^ {1 / 2} = 0, 8 0 3 * 1 0 ^ {3} \frac {m}{s} = 8 0 3 \frac {m}{s}$$

Helium gas is monoatomic gas so $E_0 = \frac{3}{2} kT = m_0\frac{V^2}{2}\rightarrow T = m_0\frac{V^2}{3k}$

$$m _ {0} = 4 * 1, 6 6 * 1 0 ^ {- 2 7} = 6, 6 4 * 1 0 ^ {- 2 7} k g$$ $$T = 6, 6 4 * 1 0 ^ {- 2 7} * \frac {2 0 ^ {2}}{3 * 1 , 3 8 * 1 0 ^ {- 2 3}} = 6 4 1, 5 * 1 0 ^ {- 4} K = 0, 0 6 4 K = - 2 7 3, 0 8 6 C$$ $$C _ {v} = \frac {\delta Q}{\delta T} = \frac {\delta U}{\delta T} \rightarrow C _ {v} (N _ {2}) = \frac {\delta (\frac {5}{2} k T)}{\delta T} = \frac {5}{2} k; C _ {v} (H e) = \frac {\delta (\frac {3}{2} k T)}{\delta T} = \frac {3}{2} k$$

Question 02:

a) Ice->water->steam:

Heat of fusion for ice: $q = 335 \, \text{kJ/kg}$ . Warmthwhichisnecessaryforicethawing: $Q = q * m = 335 * 0,2 = 67 \, \text{kJ}$ .

$$\Delta S _ {1} = \frac {Q}{T _ {0}} = \frac {6 7}{2 7 3} = 0, 2 4 5 \frac {k J}{K} = 2 4 5 \frac {J}{K}$$

Heat capacity for water: $c = 4,187 \, \text{kJ/kg}^*$ K. Warmthwhichisnecessaryforheatingwaterfrom $T_0 = 0C$ (273K) to $T = 100C$ (373 K):

$$Q = c * m * (T - T _ {0}) = 4, 1 8 7 * 0, 2 * 1 0 0 = 8 3, 7 4 k J$$ $$\Delta S _ {2} = Q * \ln \frac {T}{T _ {0}} = 83,74 * \ln \left(\frac {373}{273}\right) = 25959 \frac {J}{K}$$

Heat of vaporization for water: $L = 2260 \, \text{kJ/kg}$. Warmth which is necessary for water vaporization: $Q = Lm = 2260 * 0,2 = 452 \, \text{kJ}$.

$$\Delta S _ {3} = \frac {Q}{T} = \frac {452}{373} = 1211,8 \frac {J}{K}$$

Heat capacity for steam: $c = 2 \, \text{kJ/kg} \times \text{K}$. Warmth which is necessary for heating steam from $T = 100 \, \text{C}$ (373K) to $T_1 = 270 \, \text{C}$ (543 K):

$$Q = c * m * (T _ {1} - T) = 2 * 0,2 * 170 = 68 \, \text{kJ}$$ $$\Delta S _ {4} = Q * \ln \frac {T _ {1}}{T} = 68 * \ln \left(\frac {543}{373}\right) = 29172 \frac {J}{K}$$ $$\Delta S = 245 + 25959 + 1211,8 + 29172 = 56587,8 \frac {J}{K} - \text{entropy change for ice}.$$

Theakeloseswarmth: $Q = -(67 + 83,74 + 452 + 543) = -1145,74 \, \text{kJ}$. But temperature of the large lake is a constant so entropy change for lake: $\Delta S = \frac{Q}{T} = \frac{-1145,74}{543} = -2110 \frac{J}{K}$

The total entropy change of ice-lake system: $\Delta S = 56587,8 + (-2110) = 54477,8 \frac{J}{K}$

b) Heat of fusion for ice: $q = 335 \, \text{kJ/kg}$. Warmth which is necessary for ice thawing: $Q = q * m = 335 * 0,2 = 67 \, \text{kJ}$. $\Delta S_1 = \frac{Q}{T_0} = \frac{67}{273} = 0,245 \frac{\text{kJ}}{\text{K}} = 245 \frac{\text{J}}{\text{K}}$

Heat capacity for water: $c = 4,187 \, \text{kJ/kg} \times \text{K}$. Warmth which is necessary for heating water from $T_0 = 0 \, \text{C}$ (273K) to $T = 0,2 \, \text{C}$ (273,2 K): $Q = c * m * (T - T_0) = 4,187 * 0,2 * 0,2 = 0,167 \, \text{kJ}$

$$\Delta S _ {2} = Q * \ln \frac {T}{T _ {0}} = 0,167 * \ln \left(\frac {273,2}{273}\right) = 0,12 \, \text{J/K}$$ $$\Delta S = \Delta S _ {1} + \Delta S _ {2} = 245,12 \frac {J}{K} - \text{entropy change for ice}.$$

Theakeloseswarmth: $Q = -(67 + 0,12) = -67,12 \, \text{kJ}$. But temperature of the large lake is a constant so entropy change for lake: $\Delta S = \frac{Q}{T} = \frac{-67,12}{273,2} = -245,68 \frac{\text{J}}{\text{K}}$

The total entropy change of ice-lake system: $\Delta S = 245,12 + (-245,68) = -0,56 \frac{\text{J}}{\text{K}}$

If we don't take in attention of heating of water, the total entropy change of ice-lake system is $0 \frac{\text{J}}{\text{K}}$.

The total entropy of an isolated system always increases over time therefore it is irreversible process.

**Answer:**

Question 01: average velocity of nitrogen molecules is $803\frac{m}{s}$; temperature of Helium gas is $0,064K = -273,086C$

Question 02: The total entropy change of ice-lake system is $54477,8\frac{J}{K}$ and $0\frac{J}{K}$