Question

One kilogram of \(R-134\) a fills a \(0.090 \mathrm{m}^{3}\) rigid container at an initial temperature of \(-40^{\circ} \mathrm{C}\). The container is then heated until the pressure is 280 kPa. Determine the initial pressure and final temperature.

Short answer

Answer: The initial pressure is approximately 246.29 kPa, and the final temperature is approximately 36.39°C.

Step-by-step solution

Find the initial specific volume

To find the initial pressure, we need to first find the specific volume (volume per unit mass) at the given initial conditions. The specific volume, \(v_1\), can be calculated as follows: $$ v_1 = \frac{V}{m} $$ where \(V\) is the volume, and \(m\) is the mass. In this case, \(V = 0.090 \mathrm{m}^3\) and \(m = 1kg\). Plugging in these values, we get: $$ v_1 = \frac{0.090 \mathrm{m}^3}{1\mathrm{kg}} = 0.090 \mathrm{m}^3\mathrm{kg}^{-1} $$

Use properties of \(R-134a\) to find the initial pressure

For \(R-134a\), we can use property tables to find the specific volume at a given temperature. At the given initial temperature of \(-40^{\circ}\mathrm{C}\), we can use the saturated conditions table for \(R-134a\) to find the values of specific volume for the saturated liquid and vapor phases. Using the data from the table, we can determine the initial pressure using interpolation. First, we find the values of specific volume for the saturated liquid and vapor phases at the given initial temperature (\(T_1 = -40^{\circ}\mathrm{C}\)): $$ v_{f1} = 0.000734 \mathrm{m}^3\mathrm{kg}^{-1} $$ $$ v_{g1} = 0.05161 \mathrm{m}^3\mathrm{kg}^{-1} $$ Since our initial specific volume (\(0.090 \mathrm{m}^3\mathrm{kg}^{-1}\)) falls between these values, we know the refrigerant is in the two-phase region (a mixture of liquid and vapor). We can find the quality, \(x_1\), at the initial state using the formula: $$ x_1 = \frac{v_1 - v_{f1}}{v_{g1} - v_{f1}} $$ Plugging in the values, we get: $$ x_1 = \frac{0.090 - 0.000734}{0.05161 - 0.000734} = 1.733 $$ Since the value of \(x_1\) is greater than 1, it means that our initial assumption of being in the two-phase region was incorrect. It indicates that the refrigerant is in the superheated vapor region. In this case, we can use the ideal gas law equation to find the initial pressure \(P_1\): $$ P_1 = \frac{m R T_1}{V} $$ where \(R\) is the specific gas constant for \(R-134a\), which is approximately \(R = 0.0815 \mathrm{kJ}\cdot\mathrm{kg}^{-1}\cdot\mathrm{K}^{-1}\). Therefore, we can find the initial pressure: $$ P_1 = \frac{1\mathrm{kg} \cdot 0.0815 \mathrm{kJ}\cdot\mathrm{kg}^{-1}\cdot\mathrm{K}^{-1} \cdot (273.15 - 40) \mathrm{K}}{0.090 \mathrm{m}^3} = 246.291\,\mathrm{kPa} $$ So, the initial pressure is approximately \(246.29\,\mathrm{kPa}\).

Find the final temperature

We are given that the final pressure is \(280\,\mathrm{kPa}\). Since the container is rigid, the final specific volume \(v_2\) is the same as the initial specific volume \(v_1\): $$ v_2 = v_1 = 0.090\,\mathrm{m}^3\mathrm{kg}^{-1} $$ Now, we can use the ideal gas law equation again to find the final temperature \(T_2\): $$ T_2 = \frac{P_2 V}{m R} $$ Plugging the values, we get: $$ T_2 = \frac{280\,\mathrm{kPa} \cdot 0.090\,\mathrm{m}^3}{1\mathrm{kg} \cdot 0.0815\,\mathrm{kJ}\cdot\mathrm{kg}^{-1}\cdot\mathrm{K}^{-1}} = 309.539\,\mathrm{K} $$ Converting back to Celsius: $$ T_2 = 309.539 - 273.15 = 36.389^{\circ}\mathrm{C} $$ So, the final temperature is approximately \(36.39^{\circ}\mathrm{C}\). In conclusion, the initial pressure is \(246.29\,\mathrm{kPa}\), and the final temperature is \(36.39^{\circ}\mathrm{C}\).