Question

\(100-\mathrm{kg}\) of \(\mathrm{R}-134 \mathrm{a}\) at \(200 \mathrm{kPa}\) are contained in a pistoncylinder device whose volume is \(12.322 \mathrm{m}^{3} .\) The piston is now moved until the volume is one-half its original size. This is done such that the pressure of the \(R-134\) a does not change. Determine the final temperature and the change in the total internal energy of the \(\mathrm{R}-134\) a.

Short answer

Question: Determine the final temperature and change in total internal energy during a constant pressure process of R-134a contained in a piston-cylinder device when the initial conditions are mass 100 kg, pressure 200 kPa, and volume 12.322 m³, and the volume is halved at the end of the process. Answer: The final temperature is 160.13 K, and the change in total internal energy is -11538.71 kJ.

Step-by-step solution

Identify the Given Information and the Unknown Variables

We are given the following information: 1. Mass of R-134a: \(100 \thinspace kg\) 2. Initial pressure: \(P_1 = 200 \thinspace kPa\) 3. Initial volume: \(V_1 = 12.322 \thinspace m^3\) 4. Final volume: \(V_2 = 0.5 * V_1 = 6.161 \thinspace m^3\) (since the volume is halved) We are asked to determine the final temperature (\(T_2\)) and the change in total internal energy (\(\Delta U\)).

Applying the Ideal Gas Law

We use the Ideal Gas Law to find the initial temperature. The Ideal Gas Law is given by \(PV = mRT\) Where P is the pressure, V is the volume, m is the mass, R is the specific gas constant for R-134a, and T is the temperature. The specific gas constant for R-134a is \(R=R_u/M\), where \(R_u=8.314 \thinspace kJ/(kmol \thinspace K)\) is the universal gas constant and \(M=102 \thinspace kg/kmol\) is the molar mass of R-134a. Solving for the initial temperature (\(T_1\)), we get: \(T_1 = \frac{P_1V_1}{mR}\)

Calculate the Initial Temperature

Now, we will calculate \(T_1\): \(T_1 = \frac{200 \thinspace kPa \cdot 12.322 \thinspace m^3}{100 \thinspace kg \cdot \frac{8.314 \thinspace kJ}{102 \thinspace kg/kmol \thinspace K}}\) \(T_1 = 320.26 \thinspace K\)

Calculate the Final Temperature

Since the process occurs at constant pressure, we use \(P_1 = P_2\) and \(V_1 / T_1 = V_2 / T_2\). Now, we can find the final temperature \(T_2\): \(T_2 = \frac{V_2}{V_1}T_1\) \(T_2 = \frac{6.161 \thinspace m^3}{12.322 \thinspace m^3} \cdot 320.26 \thinspace K\) \(T_2 = 160.13 \thinspace K\)

Find the Change in Internal Energy

For R-134a, we can use the specific heat at constant volume (\(c_v\)) to find the change in total internal energy. The specific heat at constant volume for R-134a is \(c_v = 0.721 \thinspace kJ/kgK\). The change in internal energy is given by: \(\Delta U = mc_v(T_2 - T_1)\) Now, let's calculate \(\Delta U\): \(\Delta U = 100 \thinspace kg \cdot 0.721 \thinspace kJ/kgK \cdot (160.13 \thinspace K - 320.26 \thinspace K)\) \(\Delta U = -11538.71 \thinspace kJ\) The final temperature is \(160.13 \thinspace K\), and the change in total internal energy is \(-11538.71 \thinspace kJ\).