A quantity of gas occupies 0.2m³ at a pressure of 1000kN/m² and a temperature of 348°C. it is first heated at constant volume until the pressure reaches 4MPa. It is then expanded adiabatically until the pressure reaches is 280kPa. Gamma for the gas is 1.41 and R=0.228 kJ/kgK. Calculate:

The initial temperature is given:

$$T_1 = 348^\circ\text{C} = 348 + 273.15 = 621.15 \, \text{K}$$

Process 1-2 is a constant volume process ($V_1 = V_2$). For an ideal gas, the relationship between pressure and temperature at constant volume is:

$$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$

Rearrange to solve for $T_2$:

$$T_2 = T_1 \left(\frac{P_2}{P_1}\right)$$

Substitute the values:

$$T_2 = 621.15 \, \text{K} \times \left(\frac{4000 \, \text{kPa}}{1000 \, \text{kPa}}\right)$$ $$T_2 = 621.15 \, \text{K} \times 4 = 2484.6 \, \text{K}$$

Process 2-3 is an adiabatic expansion. For an adiabatic process, the relationship between temperature and pressure is:

$$T_2 P_2^{(1-\gamma)/\gamma} = T_3 P_3^{(1-\gamma)/\gamma}$$

Rearrange to solve for $T_3$:

$$T_3 = T_2 \left(\frac{P_3}{P_2}\right)^{(\gamma-1)/\gamma}$$

Substitute the values:

$$T_3 = 2484.6 \, \text{K} \times \left(\frac{280 \, \text{kPa}}{4000 \, \text{kPa}}\right)^{(1.41-1)/1.41}$$ $$T_3 = 2484.6 \, \text{K} \times \left(0.07\right)^{0.41/1.41}$$ $$T_3 = 2484.6 \, \text{K} \times \left(0.07\right)^{0.29078}$$ $$T_3 = 2484.6 \, \text{K} \times 0.5096$$ $$T_3 \approx 1266.2 \, \text{K}$$

The temperature at the beginning of the process is $\boxed{\text{621.15 K}}$.
The temperature at the end of expansion is $\boxed{\text{1266.2 K}}$.

Use the ideal gas law for state 1: $P_1 V_1 = m R T_1$.

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

Substitute the values:

$$m = \frac{1000 \, \text{kPa} \times 0.2 \, \text{m}^3}{0.228 \, \text{kJ/kgK} \times 621.15 \, \text{K}}$$ $$m = \frac{200}{141.6222} \, \text{kg}$$ $$m \approx 1.4122 \, \text{kg}$$

The relationship between $C_v$, $R$, and $\gamma$ is $C_v = \frac{R}{\gamma - 1}$.

$$C_v = \frac{0.228 \, \text{kJ/kgK}}{1.41 - 1}$$ $$C_v = \frac{0.228}{0.41} \, \text{kJ/kgK}$$ $$C_v \approx 0.5561 \, \text{kJ/kgK}$$

For a constant volume process, the heat supplied is equal to the change in internal energy:

$$Q_{1-2} = m C_v (T_2 - T_1)$$

Substitute the values:

$$Q_{1-2} = 1.4122 \, \text{kg} \times 0.5561 \, \text{kJ/kgK} \times (2484.6 \, \text{K} - 621.15 \, \text{K})$$ $$Q_{1-2} = 1.4122 \times 0.5561 \times 1863.45 \, \text{kJ}$$ $$Q_{1-2} \approx 1463.0 \, \text{kJ}$$

The heat energy supplied is $\boxed{\text{1463.0 kJ}}$.

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

$$W_{2-3} = m C_v (T_2 - T_3)$$

Substitute the values:

$$W_{2-3} = 1.4122 \, \text{kg} \times 0.5561 \, \text{kJ/kgK} \times (2484.6 \, \text{K} - 1266.2 \, \text{K})$$ $$W_{2-3} = 1.4122 \times 0.5561 \times 1218.4 \, \text{kJ}$$ $$W_{2-3} \approx 956.4 \, \text{kJ}$$

The work energy transferred is $\boxed{\text{956.4 kJ}}$.