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Asked by: manish kumar
Created at: 2025-10-02 11:45:12
Final Answer:
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Let's break down and solve each part step by step:
---
## Given Data
- **Initial Volume, \( V_1 \) = .4 m³**
- **Initial Pressure, \( P_1 \) = 100 kPa**
- **Initial Temperature, \( T_1 \) = 80°C = 353.15 K**
- **Process:** Isothermal compression (\( T_2 = T_1 \))
- **Final Volume, \( V_2 \):** Not given, but to be left as a variable for now.
We'll use the **ideal gas law** throughout.
---
a) Draw the Piston/Cylinder and Note Initial Properties
```
|---------------------|
| | <--- Movable piston
| Air |
| V1 = .4 m³ |
| P1 = 100 kPa |
| T1 = 80°C = 353.15K|
|---------------------|
```
*The piston/cylinder contains air at the specified initial state above.*
---
## b) Work Done During Isothermal Compression
**For an isothermal (constant temperature) process for an ideal gas:**
\[
W = nRT \ln\left(\frac{V_2}{V_1}\right)
\]
But since temperature is constant, and from the ideal gas law \( P_1 V_1 = P_2 V_2 \):
\[
n = \frac{P_1 V_1}{RT_1}
\]
So,
\[
W = nRT \ln\left(\frac{V_2}{V_1}\right) = P_1 V_1 \ln\left(\frac{V_2}{V_1}\right)
\]
Alternatively, since \( V_2 = \frac{P_1 V_1}{P_2} \):
\[
W = P_1 V_1 \ln\left(\frac{P_1}{P_2}\right)
\]
**Leave \( V_2 \) or \( P_2 \) as unknown for now; will be found in part (c).**
---
## c) Determine the Final Pressure (State 2)
Since the temperature is constant:
\[
P_1 V_1 = P_2 V_2
\]
So,
\[
P_2 = \frac{P_1 V_1}{V_2}
\]
If the final volume \( V_2 \) is not specified, you express \( P_2 \) in terms of \( V_2 \).
---
## d) Draw a P-V Diagram
- The **P-V graph** for an isothermal process is a **hyperbola** (since \( PV = \text{constant} \)).
- Mark **two points**:
- (V₁, P₁) = (.4 m³, 100 kPa)
- (V₂, P₂)
**Sketch:**
```
|
P | ● (V1,P1)
r | .
e | .
s | .
s | ● (V2, P2)
u | (curve: PV = const)
r |
e +-------------------------
V (Volume)
```
- The curve decreases as volume decreases and pressure increases for compression.
---
## **Summary Table**
| State | P (kPa) | V (m³) | T (K) |
|-------|---------|--------|----------|
| 1 | 100 | .4 | 353.15 |
| 2 | P₂ | V₂ | 353.15 |
---
## **Final Answers (in terms of V₂ or P₂):**
### **b) Work Done:**
\[
\boxed{W = P_1 V_1 \ln\left(\frac{V_2}{V_1}\right) = 100 \times .4 \times \ln\left(\frac{V_2}{.4}\right)~\text{kJ}}
\]
### **c) Final Pressure:**
\[
\boxed{P_2 = \frac{P_1 V_1}{V_2} = \frac{100 \times .4}{V_2}~\text{kPa}}
\]
### **d) P-V Diagram:**
- Isothermal curve from (.4 m³, 100 kPa) to (V₂, P₂), as drawn above.
---
**If you know the final volume (or pressure), you can substitute and get the numerical answers! If you have a specific final volume or pressure, let me know and I can plug in the numbers.**
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