give answer in 3 steps with explanation at the end of each step and final answer at the endProblem 1

The state of an ideal gas with Cp = (5/2)R is changed from P; = 1 bar and V{ =10
m3 to P, = 10

Question

give answer in 3 steps with explanation at the end of each step and final answer  at the endProblem 1

The state of an ideal gas with Cp = (5/2)R is changed from P; = 1 bar and V{ =10
m3 to P, = 10 bar and Vf=1 m? by the following mechanically reversible
processes:

(a) Isothermal compression.

(b) Adiabatic compression followed by cooling at constant volume.

(c) Adiabatic compression followed by cooling at constant pressure.

(d) Cooling at constant pressure followed by heating at constant volume.

(e) Heating at constant volume followed by cooling at constant pressure
Calculate Q, W, AU, and AH; for each of these processes, and neatly sketch the
paths of all processes on a single PV diagram. Show which direction the path is
going in (arrows).

VIPSolutions · Accepted Answer

Let's solve **part (a): Isothermal compression** step by step.  
We'll use the following data:
- $ C_P = \frac{5}{2}R $
- $ P_1 = 1\, \text{bar} $, $ V_1 = 10\, \text{m}^3 $
- $ P_2 = 10\, \text{bar} $, $ V_2 = 1\, \text{m}^3 $
- **Isothermal process**: $ T_1 = T_2 $

We'll calculate $ Q $ (heat), $ W $ (work), $ \Delta U_t $ (change in internal energy), $ \Delta H_t $ (change in enthalpy).

---

### **Step 1: Find the Number of Moles and Temperature**

**Equation:**  
For an ideal gas: $ PV = nRT $

At state 1:  
$ P_1 V_1 = nRT_1 $

At state 2 (since isothermal):  
$ P_2 V_2 = nRT_2 $

But since $ T_1 = T_2 $ (isothermal),  
$ P_1 V_1 = P_2 V_2 $  
Check: $ 1 \times 10 = 10 \times 1 $ → $ 10 = 10 $ so OK!

Let’s find $ nT $ (since R cancels):

$ nRT_1 = P_1 V_1 $  
Let’s keep $ nRT_1 = 10\, \text{bar} \cdot \text{m}^3 $

**Explanation:**  
We confirm the process is isothermal and can use $ P_1 V_1 = P_2 V_2 $ for an ideal gas.

---

### **Step 2: Calculate Work Done ($ W $)**

**Equation for Isothermal Work:**  
$ W = nRT \ln\left(\frac{V_2}{V_1}\right) $

Substitute $ nRT = P_1 V_1 $:  
$ W = P_1 V_1 \ln\left(\frac{V_2}{V_1}\right) $

Plug in the numbers:  
- $ P_1 = 1 \, \text{bar} = 1 \times 10^5 \, \text{Pa} $
- $ V_1 = 10 \, \text{m}^3 $
- $ V_2 = 1 \, \text{m}^3 $
- $ \ln\left(\frac{1}{10}\right) = \ln(0.1) = -2.3026 $

$ W = (1 \times 10^5) \times 10 \times (-2.3026) $  
$ W = 1 \times 10^6 \times (-2.3026) = -2.3026 \times 10^6 \, \text{J} $

**Explanation:**  
Work is negative since it's compression (work done *on* the gas).

---

### **Step 3: Calculate $ Q $, $ \Delta U_t $, and $ \Delta H_t $**

- **Isothermal:** $ \Delta U_t = 0 $ (internal energy depends only on T for ideal gas)
- **$ Q $:** By First Law, $ \Delta U = Q - W \rightarrow Q = W $ (since $ \Delta U = 0 $)
- **$ \Delta H_t $:** $ \Delta H_t = nC_P(T_2-T_1) = 0 $ (since $ T_2 = T_1 $)

So:  
- $ Q = W = -2.30 \times 10^6 \, \text{J} $
- $ \Delta U_t = 0 $
- $ \Delta H_t = 0 $

**Explanation:**  
Isothermal process for ideal gas: internal energy and enthalpy unchanged, so all work done is equal to heat transferred.

---

### **Summary Table for (a):**

| Quantity    | Value                 |
|-------------|----------------------|
| Q           | $-2.30 \times 10^6$ J |
| W           | $-2.30 \times 10^6$ J |
| $\Delta U_t$ | 0                    |
| $\Delta H_t$ | 0                    |

---

#### **PV Diagram (Sketch Instructions)**
- Draw axes: $ P $ (y) vs $ V $ (x)
- Mark initial (10, 1) and final (1, 10)
- Isothermal compression: smooth curve from (10,1) to (1,10), arrow pointing to final state.

---

**Let me know if you'd like to proceed step by step for the other processes!**

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