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 find the pressure of the system, we can use the relationship between enthalpy (ΔH), internal energy (ΔU), and the change in volume (ΔV) at constant pressure. The formula we will use is:

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

### Step-by-step Solution:

1. **Identify the given values:**
- ΔH = 50 kJ
- ΔU = 113 kJ
- ΔV = -1000 m³ (since the volume is reduced)

2. **Rearrange the enthalpy equation to solve for pressure (P):**
\[ P = \frac{\Delta H - \Delta U}{\Delta V} \]

3. **Substitute the known values into the equation:**
\[ P = \frac{50 \text{ kJ} - 113 \text{ kJ}}{-1000 \text{ m}^3} \]

4. **Calculate the numerator:**
\[ 50 \text{ kJ} - 113 \text{ kJ} = -63 \text{ kJ} \]

5. **Substitute this result back into the equation for pressure:**
\[ P = \frac{-63 \text{ kJ}}{-1000 \text{ m}^3} \]

6. **Convert kJ to kN·m (since 1 kJ = 1 kN·m):**
\[ P = \frac{63 \text{ kN·m}}{1000 \text{ m}^3} \]

7. **Perform the division:**
\[ P = 0.063 \text{ kN/m}^2 \]

8. **Convert to kN/m²:**
\[ P = 63 \text{ kN/m}^2 \]

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