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.

### Given Data:

- Mass of the gas, \( m = 12 \, \text{kg} \)
- Initial pressure, \( P_1 = 1600 \, \text{kPa} \)
- Initial volume, \( V_1 = 2 \, \text{m}^3 \)
- Final pressure, \( P_2 = 20 \, \text{kPa} \)
- The relationship \( pV^{1.2} = \text{constant} \)
- The specific internal energy of the gas decreases by \( 80 \, \text{kJ/kg} \)

We need to determine the heat transfer \( Q \) and its direction.

### Step 1: Apply the given \( pV^{1.2} = \text{constant} \) relationship
Since \( pV^{1.2} \) is constant, we can write:

\[
P_1 V_1^{1.2} = P_2 V_2^{1.2}
\]

Solving for \( V_2 \) (the final volume):

\[
V_2^{1.2} = \frac{P_1 V_1^{1.2}}{P_2}
\]

Substituting the values:

\[
V_2^{1.2} = \frac{1600 \times 2^{1.2}}{20}
\]

First, calculate \( 2^{1.2} \):

\[
2^{1.2} \approx 2.297
\]

Now,

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

Now, take the \( 1/1.2 \)-th root to find \( V_2 \):

\[
V_2 = 183.76^{\frac{1}{1.2}} \approx 58.37 \, \text{m}^3
\]

### Step 2: Apply the First Law of Thermodynamics

The first law of thermodynamics is:

\[
Q = \Delta U + W
\]

Where:
- \( Q \) is the heat transfer (what we are trying to find),
- \( \Delta U \) is the change in internal energy,
- \( W \) is the work done by the gas.

### Step 3: Calculate the Change in Internal Energy

The specific internal energy decrease is \( 80 \, \text{kJ/kg} \), and since the mass of the gas is \( 12 \, \text{kg} \), the total change in internal energy is:

\[
\Delta U = m \times (\text{change in specific internal energy}) = 12 \times (-80) = -960 \, \text{kJ}
\]

### Step 4: Calculate the Work Done

For a polytropic process, the work done \( W \) is given by:

\[
W = \frac{P_2 V_2 - P_1 V_1}{1 - n}
\]

Where \( n = 1.2 \) is the polytropic index.

Substitute the known values:

\[
W = \frac{(20 \times 58.37) - (1600 \times 2)}{1 - 1.2}
\]

Convert the pressures to \( \text{kPa} \cdot \text{m}^3 \) and divide by \( 1000 \) to get the result in kilojoules:

\[
W = \frac{(1167.4) - (3200)}{-0.2} = \frac{-2032.6}{-0.2} = 10163 \, \text{kJ}
\]

### Step 5: Calculate the Heat Transfer

Now, substitute \( \Delta U \) and \( W \) into the first law equation:

\[
Q = \Delta U + W = (-960) + 10163 = 9203 \, \text{kJ}
\]

### Step 6: Determine the Direction of Heat Transfer

Since the heat transfer \( Q = 9203 \, \text{kJ} \) is positive, it indicates that heat is **added to the system**.

### Final Answer:

The magnitude of the heat transfer is **9203 kJ**, and the direction is **into the system**.

---

Would you like more details or have any questions?

Here are 5 related questions you might find helpful:
1. What is the significance of the polytropic index in thermodynamic processes?
2. How does the specific internal energy change affect the overall energy balance?
3. What is the physical meaning of a positive heat transfer in thermodynamics?
4. How can we derive the work done for different types of thermodynamic processes?
5. How does the first law of thermodynamics apply to closed systems with varying pressures?

**Tip**: Understanding polytropic processes is crucial for analyzing real-world gas expansion and compression in thermodynamic systems!

A mass of 12kg gas expands within a flexible container so that pv relationship is of the form pV^1.2 = constant. The Initial pressure is 1600kPa and the initial volume is 2 m^3. 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.

Find the heat transfer for a gas expansion where pV^1.2 = constant, with initial conditions of 1600kPa and 2 m^3. This detailed thermodynamic solution involves applying the first law of thermodynamics, calculating internal energy change, and work done during the process. Specific internal energy decreases by 80 kJ/kg, and final heat transfer magnitude is 9203 kJ into the system.

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