Question

Evaluate and compare the work and heat transfer when 0.8kg of air, at a pressure of 2.6 Bar and a temperature of C degree, expand in a closed thermodynamic system to three times its initial volume : (1) according to ti Boyles law and, (2) according to Charles law. The characteristic gas constant for air is 200 J kg -1 K-1 and its specific heat capacity at constant volume is 700Jkg -1 K-1

Answer 1

Answer on Question#78799 - Physics - Molecular Physics

Evaluate and compare the work and heat transfer when $m = 0.8 \, \mathrm{kg}$ of air, at a pressure of $P = 2.6 \, \mathrm{Bar}$ and a temperature of $C$ degree, expand in a closed thermodynamic system to three times its initial volume: (1) according to Boyles law and, (2) according to Charles law.

The characteristic gas constant for air is $R = 200 \, \mathrm{J/kg \cdot K}$ and its specific heat capacity at constant volume is $C_V = 700 \, \mathrm{J/kg \cdot K}$.

Solution:

$T_{i} = 23{}^{\circ}\mathrm{C} = 300\mathrm{K}$ – initial temperature.

(1) According to Boyles law $PV = \text{const}$. Thus the internal energy of the gas doesn't change and the heat transfer is given by the work of the gas, which in this case is

$$\Delta Q _ {B} = m R T \ln \frac {V _ {f}}{V _ {i}},$$

where $V_{i}$ – initial volume of the gas, $V_{f}$ – final volume.

Since $V_{f} = 3V_{i}$, we obtain

$$\Delta Q _ {B} = 0. 8 \mathrm {k g} \cdot 2 0 0 \frac {\mathrm {J}}{\mathrm {k g} \cdot \mathrm {K}} \cdot 3 0 0 \mathrm {K} \ln 3 = 5 2. 7 \mathrm {k J}$$

(2) According to Charles law $V / T = \text{const}$, thus

$$\frac {V _ {i}}{T _ {i}} = \frac {V _ {f}}{T _ {f}}$$

Since $V_{f} = 3V_{i}, T_{f} = 3T_{i}$. The heat transfer is given by

$$\Delta Q _ {C} = P \Delta V + m C _ {V} \Delta T$$

According to the ideal gas law $PV = mRT$, thus the first member (the work of the gas) in the upper equation can be rewritten in the following way:

$$A _ {C} = P \Delta V = m R \Delta T = 0. 8 \mathrm {k g} \cdot 2 0 0 \frac {\mathrm {J}}{\mathrm {k g} \cdot \mathrm {K}} \cdot 2 \cdot 3 0 0 \mathrm {K} = 9 6 \mathrm {k J}$$

Therefore we obtain

$$\begin{array}{l} \Delta Q _ {C} = m R \Delta T + m C _ {V} \Delta T = m (R + C _ {V}) \Delta T = m (R + C _ {V}) 2 T _ {i} = \\ = 0. 8 \mathrm {k g} \left(2 0 0 \frac {\mathrm {J}}{\mathrm {k g} \cdot \mathrm {K}} + 7 0 0 \frac {\mathrm {J}}{\mathrm {k g} \cdot \mathrm {K}}\right) 2 \cdot 3 0 0 \mathrm {K} = 4 3 2 \mathrm {k J} \\ \end{array}$$

Thus the work of the gas is $A_{C} / \Delta Q_{B} = 96 \, \mathrm{kJ} / 52.7 \, \mathrm{kJ} = 1.8$ times greater according to Charles law, than to Boyles law. Also the heat transfer in case of Charles law is 8.2 times greater than in case of Boyles law.

Answer:

(1) $\Delta Q_{B} = 52.7\mathrm{kJ}$

$$A _ {B} = \Delta Q _ {B} = 5 2. 7 \mathrm {k J}$$

(2) $\Delta Q_{C} = 432\mathrm{kJ}$

$$A _ {C} = 9 6 \mathrm {k J}$$

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