To determine the pressure of the gas when the volume changes while keeping the temperature constant, we can use Boyle's Law. This law states that for a given mass of an ideal gas at a constant temperature, the pressure of the gas is inversely proportional to its volume. In mathematical terms, Boyle's Law is given by:
[tex]\[ P_1 \cdot V_1 = P_2 \cdot V_2 \][/tex]
Where:
- [tex]\( P_1 \)[/tex] is the initial pressure of the gas.
- [tex]\( V_1 \)[/tex] is the initial volume of the gas.
- [tex]\( P_2 \)[/tex] is the final pressure of the gas.
- [tex]\( V_2 \)[/tex] is the final volume of the gas.
Let's go through the problem step-by-step:
### Formula:
[tex]\[ P_1 \cdot V_1 = P_2 \cdot V_2 \][/tex]
### Input Numbers:
- Initial pressure ([tex]\( P_1 \)[/tex]): 15 Pa
- Initial volume ([tex]\( V_1 \)[/tex]): 5 mL
- Final volume ([tex]\( V_2 \)[/tex]): 2 mL
- Temperature ([tex]\( T \)[/tex]): 3 K (constant, and thus irrelevant for Boyle's Law)
### Solve for the final pressure ([tex]\( P_2 \)[/tex]):
1. Rearrange the formula to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{P_1 \cdot V_1}{V_2} \][/tex]
2. Substitute the given values into the formula:
[tex]\[ P_2 = \frac{15 \, \text{Pa} \cdot 5 \, \text{mL}}{2 \, \text{mL}} \][/tex]
3. Calculate the result:
[tex]\[ P_2 = \frac{75 \, \text{Pa} \cdot \text{mL}}{2 \, \text{mL}} = 37.5 \, \text{Pa} \][/tex]
### Answer:
The pressure would be 37.5 Pa if the volume shrinks to 2 mL while the temperature remains at 3 K.
