Explain Internal energy of gas. Draw P-V diagram indicate adiabatic expansion process show initial and final stage.

The relationship between $C_p$, $C_v$, and $R$ is given by:

$$C_p = C_v + R$$

Substitute the given values:

$$C_p = 0.312 \, \text{kJ/kgK} + 0.208 \, \text{kJ/kgK}$$ $$C_p = 0.520 \, \text{kJ/kgK}$$

The adiabatic index (or index of compression for an adiabatic process) is given by:

$$\gamma = \frac{C_p}{C_v}$$

Substitute the calculated $C_p$ and given $C_v$:

$$\gamma = \frac{0.520 \, \text{kJ/kgK}}{0.312 \, \text{kJ/kgK}}$$ $$\gamma = 1.6666...$$

Rounding to three decimal places:

$$\gamma \approx 1.667$$

The value of the index of compression for the adiabatic process is $\boxed{\text{1.667}}$.

1.3.2. The all unknown pressures

The ideal gas law is $PV = mRT$. For the initial state:

$$P_1 V_1 = m R T_1$$ $$P_1 = \frac{m R T_1}{V_1}$$

Substitute the given values:

$$P_1 = \frac{4.15 \, \text{kg} \times 0.208 \, \text{kJ/kgK} \times 317.15 \, \text{K}}{1.51 \, \text{m}^3}$$ $$P_1 = \frac{4.15 \times 0.208 \times 317.15}{1.51} \, \text{kPa}$$ $$P_1 = \frac{273.8696}{1.51} \, \text{kPa}$$ $$P_1 \approx 181.37 \, \text{kPa}$$

Rearrange the formula to solve for $V_2$:

$$\frac{T_1}{T_2} = \left(\frac{V_2}{V_1}\right)^{\gamma-1}$$ $$V_2 = V_1 \left(\frac{T_1}{T_2}\right)^{\frac{1}{\gamma-1}}$$

Substitute the values:

$$V_2 = 1.51 \, \text{m}^3 \left(\frac{317.15 \, \text{K}}{435.15 \, \text{K}}\right)^{\frac{1}{1.667-1}}$$ $$V_2 = 1.51 \left(\frac{317.15}{435.15}\right)^{\frac{1}{0.667}}$$ $$V_2 = 1.51 \left(0.7288\right)^{1.499}$$ $$V_2 = 1.51 \times 0.620$$ $$V_2 \approx 0.936 \, \text{m}^3$$

$$P_2 V_2 = m R T_2$$ $$P_2 = \frac{m R T_2}{V_2}$$

Substitute the values:

$$P_2 = \frac{4.15 \, \text{kg} \times 0.208 \, \text{kJ/kgK} \times 435.15 \, \text{K}}{0.936 \, \text{m}^3}$$ $$P_2 = \frac{4.15 \times 0.208 \times 435.15}{0.936} \, \text{kPa}$$ $$P_2 = \frac{375.69}{0.936} \, \text{kPa}$$ $$P_2 \approx 401.38 \, \text{kPa}$$

The unknown pressures are $P_1 = \boxed{\text{181.37 kPa}}$ and $P_2 = \boxed{\text{401.38 kPa}}$.

1.3.3. The change in internal energy of the gas during the process

The change in internal energy for a gas is given by:

$$\Delta U = m C_v (T_2 - T_1)$$

Substitute the given values:

$$\Delta U = 4.15 \, \text{kg} \times 0.312 \, \text{kJ/kgK} \times (435.15 \, \text{K} - 317.15 \, \text{K})$$ $$\Delta U = 4.15 \times 0.312 \times 118 \, \text{kJ}$$ $$\Delta U = 152.5056 \, \text{kJ}$$

Rounding to two decimal places:

$$\Delta U \approx 152.51 \, \text{kJ}$$

The change in internal energy of the gas is $\boxed{\text{152.51 kJ}}$.

1.3.4. The work done during compression.

For an adiabatic process, the work done is given by:

$$W = \frac{P_2 V_2 - P_1 V_1}{1 - \gamma}$$

Substitute the calculated pressures and volumes, and the adiabatic index:

$$W = \frac{(401.38 \, \text{kPa} \times 0.936 \, \text{m}^3) - (181.37 \, \text{kPa} \times 1.51 \, \text{m}^3)}{1 - 1.667}$$ $$W = \frac{375.70 - 273.87}{-0.667} \, \text{kJ}$$ $$W = \frac{101.83}{-0.667} \, \text{kJ}$$ $$W \approx -152.67 \, \text{kJ}$$

The negative sign indicates work is done on the gas (compression).
The work done during compression is $\boxed{\text{-152.67 kJ}}$.