Internal Energy (E,also denoted by U):
Every system having some quantity of matter is associated with a definite amount of energy, called internal energy.
`E=E_("Transiational")+E_("Rotational")+E_("Vibrational")+E_("bonding")+`....
`DeltaE=E_("Final")-E_("initial")`
`DeltaE=q_v`, heat supplied to a gas at constant volume, since all the heat supplied goes to increase the internal energy to the gas.
It is an extensive property & a state function.It is exclusively a function of temperature.
If `DeltaT=0 , DeltaE=0` as well.
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.
If the expansion is quasi-static, then the value of q is :

Options

A. 80 kJ
B. 60 kJ
C. 40 kJ
D. 120 kJ

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the initial and final states
- Initial pressure \( P_1 = 400 \, \text{kPa} \)
- Initial volume \( V_1 = 0.2 \, \text{m}^3 \)
- Final pressure \( P_2 = 100 \, \text{kPa} \)

### Step 2: Use the relation \( PV^2 = \text{constant} \) to find the final volume \( V_2 \)
Using the relation:
\[ P_1 V_1^2 = P_2 V_2^2 \]
Substituting the known values:
\[ 400 \times (0.2)^2 = 100 \times V_2^2 \]
Calculating \( 400 \times 0.04 = 16 \):
\[ 16 = 100 \times V_2^2 \]
Solving for \( V_2^2 \):
\[ V_2^2 = \frac{16}{100} = 0.16 \]
Taking the square root:
\[ V_2 = \sqrt{0.16} = 0.4 \, \text{m}^3 \]

### Step 3: Calculate the change in internal energy \( \Delta U \)
The internal energy is given by:
\[ U = 3PV + 84 \]
Calculating \( U \) for the final state:
\[ U_2 = 3 \times P_2 \times V_2 + 84 = 3 \times 100 \times 0.4 + 84 = 120 + 84 = 204 \, \text{kJ/kg} \]
Calculating \( U \) for the initial state:
\[ U_1 = 3 \times P_1 \times V_1 + 84 = 3 \times 400 \times 0.2 + 84 = 240 + 84 = 324 \, \text{kJ/kg} \]
Now, calculate \( \Delta U \):
\[ \Delta U = U_2 - U_1 = 204 - 324 = -120 \, \text{kJ} \]

### Step 4: Calculate the work done \( W \)
For a polytropic process where \( n = 2 \):
\[ W = \frac{P_2 V_2 - P_1 V_1}{n - 1} \]
Substituting the values:
\[ W = \frac{100 \times 0.4 - 400 \times 0.2}{2 - 1} = \frac{40 - 80}{1} = -40 \, \text{kJ} \]

### Step 5: Apply the first law of thermodynamics to find \( Q \)
The first law states:
\[ \Delta U = Q + W \]
Rearranging gives:
\[ Q = \Delta U - W \]
Substituting the values:
\[ Q = -120 - (-40) = -120 + 40 = -80 \, \text{kJ} \]

### Final Answer
The value of \( Q \) is:
\[ \boxed{-80 \, \text{kJ}} \]