A mass of 12kg gas expands within a flexible container so that pv relationship is of the form pvl. 2= constant.
The Initial pressure is 1600kPa and the initial volume is 2 m3.The final pressure is 20 kPa. If specific internal energy of the gas decreases by 80kJ/kg.Find the heat transfer in magnitude and direction.


To solve this, let's break down the information and use the first law of thermodynamics for a closed system. Given:

- **Initial pressure, \( P_1 = 1600 \, \text{kPa} \)**
- **Initial volume, \( V_1 = 2 \, \text{m}^3 \)**
- **Final pressure, \( P_2 = 20 \, \text{kPa} \)**
- **Mass of gas, \( m = 12 \, \text{kg} \)**
- **Specific internal energy change, \( \Delta u = -80 \, \text{kJ/kg} \)** (decrease, so it's negative)

We’re asked to find the heat transfer, \( Q \).

### Step 1: Determine the Final Volume \( V_2 \)
The given relation is:
\[
P V^{1.2} = \text{constant}
\]

Using the initial state:
\[
P_1 V_1^{1.2} = P_2 V_2^{1.2}
\]
Substitute the known values:
\[
1600 \times 2^{1.2} = 20 \times V_2^{1.2}
\]

First, calculate \( 2^{1.2} \):
\[
2^{1.2} \approx 2.297
\]

Then substitute:
\[
1600 \times 2.297 = 20 \times V_2^{1.2}
\]
\[
V_2^{1.2} = \frac{1600 \times 2.297}{20} = 183.76
\]

Now, solve for \( V_2 \):
\[
V_2 = (183.76)^{\frac{1}{1.2}} \approx 80.18 \, \text{m}^3
\]

### Step 2: Calculate Work Done \( W \) Using the \( PV \) Relationship
The work done \( W \) for a polytropic process \( PV^n = \text{constant} \) is given by:
\[
W = \frac{P_2 V_2 - P_1 V_1}{1 - n}
\]
where \( n = 1.2 \).

Substitute the values:
\[
W = \frac{(20 \times 80.18) - (1600 \times 2)}{1 - 1.2}
\]
Calculate each term:
\[
P_2 V_2 = 20 \times 80.18 = 1603.6 \, \text{kJ}
\]
\[
P_1 V_1 = 1600 \times 2 = 3200 \, \text{kJ}
\]

Thus:
\[
W = \frac{1603.6 - 3200}{-0.2} = \frac{-1596.4}{-0.2} = 7982 \, \text{kJ}
\]

### Step 3: Apply the First Law of Thermodynamics to Find \( Q \)
The first law of thermodynamics states:
\[
Q = \Delta U + W
\]
where \( \Delta U \) is the change in internal energy and is given by:
\[
\Delta U = m \Delta u = 12 \times (-80) = -960 \, \text{kJ}
\]

Now substitute values into the first law:
\[
Q = -960 + 7982 = 7022 \, \text{kJ}
\]

### Conclusion
The heat transfer \( Q \) is **7022 kJ**. Since \( Q \) is positive, heat flows **into the system**.

---

Would you like further details on any steps, or do you have any questions?

---

1. How would the solution change if the final pressure was different?
2. What would happen if the specific internal energy increased instead of decreased?
3. How does the polytropic exponent affect work done calculations?
4. How does this process compare with an isothermal process in terms of work done?
5. How is heat transfer affected by mass in closed systems?

**Tip:** In polytropic processes, knowing the polytropic index \( n \) can help determine how pressure and volume interact during changes.

A mass of 12kg gas expands within a flexible container so that pv relationship is of the form pvl. 2= constant. The Initial pressure is 1600kPa and the initial volume is 2 m3. The final pressure is 20 kPa. If specific internal energy of the gas decreases by 80kJ/kg. Find the heat transfer in magnitude and direction.

Explore how to calculate heat transfer for a gas expanding in a polytropic process, given initial and final pressures, volume, mass, and specific internal energy change. This problem uses thermodynamic principles, including the First Law of Thermodynamics, and is suited for advanced high school or college-level physics students.

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