Consider 1 kg of an ideal gas at 1 bar and 300 K contained in a rigid and perfectly insulated container. The specific heat of the gas at constant volume c_v is equal to 750 J-kg^-1K^-1. A stirrer performs 225 kJ of work on the gas. Assume that the container does not participate in the thermodynamic interaction. The final pressure of the gas will be ______ bar (in integer).

The correct answer is 2.

As shown in the schematic diagram of the container, an ideal gas is kept inside a rigid, perfectly insulated vessel with a stirrer mechanism.

According to the First Law of Thermodynamics:
$$dQ=dU+dW$$
Here:
1. The container is perfectly insulated, meaning there is no heat transfer:
$dQ=0$
2. The container is rigid, which means there is no boundary or displacement work:
$d{W}_{\text{disp}}=0$
3. The only work done is the shaft work performed by the stirrer on the gas. Work done on the gas is negative:
$d{W}_{\text{stirrer}}=-225\text{kJ}=-225000\text{J}$

Substituting these values into the first law equation:
$$0=dU+d{W}_{\text{stirrer}}$$
$$dU=-d{W}_{\text{stirrer}}=-(-225000\text{J})=225000\text{J}$$

The change in internal energy for an ideal gas undergoing a constant volume process is related to the specific heat capacity at constant volume _v by:
$$dU=m{c}_v(T_2-T_1)$$

Given parameters from the problem description:
Mass of the gas, $m=1\text{kg}$
Initial temperature, $T_1=300\text{K}$
Specific heat, $c_v=750{\text{J kg}}^{-1}{\text{K}}^{-1}$

Substituting these values:
$$1\times 750\times (T_2-300)=225000$$
$$T_2-300=\frac{225000}{750}$$
$$T_2-300=300$$
$$T_2=600\text{K}$$

Because the process takes place in a rigid container, the volume of the gas remains constant. For an ideal gas at constant volume, pressure is directly proportional to its absolute temperature:
$$\frac{P_2}{P_1}=\frac{T_2}{T_1}$$

Substituting the initial pressure $P_1=1\text{bar}$ and the temperatures:
$$P_2=1\text{bar}\times \frac{600}{300}=2\text{bar}$$

Therefore, the final pressure of the gas is 2 bar.