To solve this problem, we'll use the Ideal Gas Law, which can be applied under the assumption that the amount of gas and the temperature remain constant. The general form of the Ideal Gas Law is:
\[ PV = nRT \]
Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles of gas (constant, since we are not adding or removing gas)
- \( R \) = Ideal gas constant
- \( T \) = Temperature in Kelvin
Step 1: Given Values
-
Initial conditions:
- \( P_1 = 0.15 , \text{MPa} = 150 , \text{kPa} \) (1 MPa = 1000 kPa)
- \( V_1 = 39 , m^3 \)
- \( T_1 = 32 , °C = 32 + 273.15 = 305.15 , K \)
-
Final conditions:
- \( P_2 = 5 , kPa \)
- \( V_2 = ? \)
- \( T_2 = T_1 = 305.15 , K \)
Step 2: Applying Boyle's Law
Since the temperature is constant, we can apply Boyle's Law, which states that for a given amount of gas at constant temperature, the product of pressure and volume is constant:
\[ P_1 V_1 = P_2 V_2 \]
Step 3: Substitute the Known Values
Now we can substitute the values into Boyle's Law equation:
\[ 150 , kPa \cdot 39 , m^3 = 5 , kPa \cdot V_2 \]
Step 4: Solve for \( V_2 \)
Rearranging the equation to solve for \( V_2 \):
\[ V_2 = \frac{150 , kPa \cdot 39 , m^3}{5 , kPa} \]
Calculating \( V_2 \):
\[ V_2 = \frac{5850 , kPa \cdot m^3}{5 , kPa} = 1170 , m^3 \]
Conclusion
The volume of the gas when the pressure decreases to 5 kPa at a constant temperature of 32 °C is:
\[ \boxed{1170 , m^3} \]