The internal energy of a certain substance is given by the following equation :
U=3 PV+84
where U is given in kJ/kg, P is in kPa, and V is in `m^3//kg`
A system composed of 3 kg of this substance expands from an initial pressure of 400 kPa and a volume of A `0.2 m^3` to a final pressure 100 kPa in a process in which pressure and volume are related by `PV^2`=constant.
In another process the same system expands according to the same pressure-volume relationship as in above question, but from the same initial state of the final state as in above question, but the heat transfer in this case is +30 kJ.Then the work transfer for this process is :

Options

A. `-80 kJ`
B. `-60 kJ`
C. `-90 kJ`
D. `-150 kJ`

Explanation

To solve the problem step by step, we will use the first law of thermodynamics and the given equations.

### Step 1: Understand the Internal Energy Equation
The internal energy \( U \) of the substance is given by the equation:
\[ U = 3PV + 84 \]
where \( U \) is in kJ/kg, \( P \) is in kPa, and \( V \) is in m³/kg.

### Step 2: Calculate Initial Internal Energy
The initial state has:
- Initial pressure \( P_i = 400 \) kPa
- Initial volume \( V_i = 0.2 \) m³/kg

Using the internal energy equation:
\[ U_i = 3 \times P_i \times V_i + 84 \]
Substituting the values:
\[ U_i = 3 \times 400 \times 0.2 + 84 \]
\[ U_i = 240 + 84 = 324 \text{ kJ/kg} \]

### Step 3: Calculate Final Volume
Given the relationship \( PV^2 = \text{constant} \), we can find the final volume \( V_f \) at the final pressure \( P_f = 100 \) kPa.

Using the initial state:
\[ 400 \times (0.2)^2 = 100 \times V_f^2 \]
Calculating the left side:
\[ 400 \times 0.04 = 16 \]
So,
\[ 16 = 100 \times V_f^2 \]
\[ V_f^2 = \frac{16}{100} = 0.16 \]
\[ V_f = \sqrt{0.16} = 0.4 \text{ m³/kg} \]

### Step 4: Calculate Final Internal Energy
Now we can calculate the final internal energy \( U_f \):
\[ U_f = 3 \times P_f \times V_f + 84 \]
Substituting the values:
\[ U_f = 3 \times 100 \times 0.4 + 84 \]
\[ U_f = 120 + 84 = 204 \text{ kJ/kg} \]

### Step 5: Calculate Change in Internal Energy
The change in internal energy \( \Delta U \) for the 3 kg of substance is:
\[ \Delta U = U_f - U_i = 204 - 324 = -120 \text{ kJ/kg} \]
For 3 kg:
\[ \Delta U_{total} = 3 \times (-120) = -360 \text{ kJ} \]

### Step 6: Apply the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ Q = \Delta U + W \]
Where:
- \( Q = 30 \) kJ (heat transfer)
- \( \Delta U = -360 \) kJ
- \( W \) is the work done by the system.

Rearranging the equation:
\[ W = Q - \Delta U \]
Substituting the values:
\[ W = 30 - (-360) = 30 + 360 = 390 \text{ kJ} \]

### Final Answer
The work transfer for this process is:
\[ W = 390 \text{ kJ} \]