1. Consider the piston cylinder assembly shown below. It is well insulated and initially contains two 5000 kg blocks at rest on the 0.05 m² piston. The initial temperature is 500 K. The ambient pressure is 5 bar. 1 mol of an ideal gas is contained in the cylinder. This gas is compressed in a process where another 5000 kg block is added. The heat capacity of the gas at constant volume can be taken to have a constant value of 5/2 R, where R is the gas constant. A) What is the initial and final pressure of the gas in the system? Do you expect the temperature to rise or fall? Explain B) What is the final temperature? Calculate Assys and Assurr. C) Does this process violate the second law of thermodynamics? Explain. m=5000 kg Process consists of adding third 5000 kg block to compress piston

Elements Of Electromagnetics
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Please Label variables and note when steam table values are used.

**Problem Statement:**

1. Consider the piston-cylinder assembly shown below. It is well-insulated and initially contains two 5000 kg blocks at rest on the 0.05 m² piston. The initial temperature is 500 K. The ambient pressure is 5 bar. 1 mol of an ideal gas is contained in the cylinder. This gas is compressed in a process where another 5000 kg block is added. The heat capacity of the gas at constant volume can be taken to have a constant value of 5/2 R, where R is the gas constant.

   **Questions:**
   A) What is the initial and final pressure of the gas in the system?
      Do you expect the temperature to rise or fall? Explain.
   
   B) What is the final temperature?
      Calculate ΔS<sub>sys</sub> and ΔS<sub>surr</sub>.
   
   C) Does this process violate the second law of thermodynamics? Explain.

**Diagram Explanation:**

The diagram depicts a piston-cylinder setup:

- It shows a well-insulated piston.
- Initially, there are two 5000 kg blocks resting on the piston.
- Another 5000 kg block is added to compress the piston further.
- P<sub>E</sub> (External Pressure) = 5 bar.
- Piston area (A) = 0.05 m².
- Contains 1 mole of ideal gas with an initial temperature (T<sub>initial</sub>) of 500 K.
- Heat capacity, C<sub>v</sub> = 5/2 R.

**Entropy Calculation Hint:**

The change in entropy (ΔS<sub>sys</sub>) for an ideal gas system with constant C<sub>v</sub>, transitioning from (T<sub>1</sub>, V<sub>1</sub>) to (T<sub>2</sub>, V<sub>2</sub>), can be calculated using the formula:

\[
\Delta S_{\text{sys}} = C_{v} \ln \left( \frac{T_2}{T_1} \right) + R \ln \left( \frac{V_2}{V_1} \right) = (5/2R) \ln \left( \frac{T_2}{T_1} \right) + R \ln \left( \
Transcribed Image Text:**Problem Statement:** 1. Consider the piston-cylinder assembly shown below. It is well-insulated and initially contains two 5000 kg blocks at rest on the 0.05 m² piston. The initial temperature is 500 K. The ambient pressure is 5 bar. 1 mol of an ideal gas is contained in the cylinder. This gas is compressed in a process where another 5000 kg block is added. The heat capacity of the gas at constant volume can be taken to have a constant value of 5/2 R, where R is the gas constant. **Questions:** A) What is the initial and final pressure of the gas in the system? Do you expect the temperature to rise or fall? Explain. B) What is the final temperature? Calculate ΔS<sub>sys</sub> and ΔS<sub>surr</sub>. C) Does this process violate the second law of thermodynamics? Explain. **Diagram Explanation:** The diagram depicts a piston-cylinder setup: - It shows a well-insulated piston. - Initially, there are two 5000 kg blocks resting on the piston. - Another 5000 kg block is added to compress the piston further. - P<sub>E</sub> (External Pressure) = 5 bar. - Piston area (A) = 0.05 m². - Contains 1 mole of ideal gas with an initial temperature (T<sub>initial</sub>) of 500 K. - Heat capacity, C<sub>v</sub> = 5/2 R. **Entropy Calculation Hint:** The change in entropy (ΔS<sub>sys</sub>) for an ideal gas system with constant C<sub>v</sub>, transitioning from (T<sub>1</sub>, V<sub>1</sub>) to (T<sub>2</sub>, V<sub>2</sub>), can be calculated using the formula: \[ \Delta S_{\text{sys}} = C_{v} \ln \left( \frac{T_2}{T_1} \right) + R \ln \left( \frac{V_2}{V_1} \right) = (5/2R) \ln \left( \frac{T_2}{T_1} \right) + R \ln \left( \
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