At T = 10°C, v = 70 m³/kg, find p in kPa and h in kJ/kg.

Steps to Solve

1. Identify the substance

   Since the substance isn't specified, we'll assume it is water. This is a common assumption in thermodynamics problems unless otherwise specified.

2. Determine the phase

   At $T = 10^\circ\text{C}$, the saturation pressure of water ($P_{sat}$) is approximately $1.2281 \text{ kPa}$. The specific volume of saturated vapor ($v_g$) at $10^\circ\text{C}$ is $57.79 \text{ m}^3/\text{kg}$. Since the given specific volume $v = 70 \text{ m}^3/\text{kg}$ is greater than $v_g$, the water is in the superheated vapor phase.

3. Look up properties in the superheated water table

   Look in the superheated water tables for $T = 10^\circ\text{C}$ and $v = 70 \text{ m}^3/\text{kg}$. Since $10^\circ\text{C}$ isn't a standard temperature in superheated tables, and the specific volume is quite high indicating very low pressure, we can approximate the behavior as an ideal gas.

4. Approximate using ideal gas law

   For water vapor at low pressures, we can use the ideal gas law: $Pv = RT$, where $R$ is the specific gas constant for water vapor.

   $R = \frac{\bar{R}}{M}$, where $\bar{R} = 8.314 \text{ kJ/kmol.K}$ is the universal gas constant and $M = 18.015 \text{ kg/kmol}$ is the molar mass of water.

   $R = \frac{8.314}{18.015} = 0.4615 \text{ kJ/kg.K}$

   Convert the temperature to Kelvin: $T = 10 + 273.15 = 283.15 \text{ K}$

   Now, we can find the pressure: $P = \frac{RT}{v} = \frac{0.4615 \text{ kJ/kg.K} \times 283.15 \text{ K}}{70 \text{ m}^3/\text{kg}} = \frac{130.68 \text{ kJ/kg}}{70 \text{ m}^3/\text{kg}} = 1.867 \text{ kPa}$

5. Determine the enthalpy

   For an ideal gas, enthalpy is a function of temperature only. At low pressures, we can approximate the enthalpy using the saturated vapor enthalpy at the given temperature. From the saturated water tables at $T = 10^\circ\text{C}$, $h_g = 2519.2 \text{ kJ/kg}$. Since the pressure is very low, we can assume the enthalpy is close to this value.