The enthalpy of a system increases by 50 kJ when its internal energy is increased by 113 kJ. What is the pressure in `k Nm^(-2)` of the system if the volume of gas is reduced by `10^(3) m^(3)` at constant pressure?

Explanation

To solve the problem, we will use the relationship between the change in enthalpy (ΔH), the change in internal energy (ΔU), and the change in volume (ΔV) at constant pressure. The equation we will use is:

\[ \Delta H = \Delta U + P \Delta V \]

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Change in enthalpy, ΔH = 50 kJ
- Change in internal energy, ΔU = 113 kJ
- Change in volume, ΔV = -1000 m³ (since the volume is reduced)

2. **Convert the Units:**
- Convert kJ to J for consistency:
- ΔH = 50 kJ = 50,000 J
- ΔU = 113 kJ = 113,000 J

3. **Substitute the Values into the Equation:**
- Using the equation ΔH = ΔU + PΔV:
\[ 50,000 = 113,000 + P(-1000) \]

4. **Rearranging the Equation:**
- Rearranging gives:
\[ P(-1000) = 50,000 - 113,000 \]
\[ P(-1000) = -63,000 \]

5. **Solve for Pressure (P):**
- Divide both sides by -1000:
\[ P = \frac{-63,000}{-1000} = 63 \text{ kN/m}^2 \]

### Final Answer:
The pressure of the system is **63 kN/m²**.