Answer :
To solve this problem, we will use Boyle's Law. Boyle's Law states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with the pressure. Mathematically, it is expressed as:
[tex]\[ P_1 \times V_1 = P_2 \times V_2 \][/tex]
where:
- [tex]\( P_1 \)[/tex] is the initial pressure
- [tex]\( V_1 \)[/tex] is the initial volume
- [tex]\( P_2 \)[/tex] is the new (final) pressure
- [tex]\( V_2 \)[/tex] is the new (final) volume
Given the values:
- [tex]\( P_1 = 4.0 \text{ atm} \)[/tex]
- [tex]\( V_1 = 275 \text{ mL} \)[/tex]
- [tex]\( V_2 = 175 \text{ mL} \)[/tex]
We need to find [tex]\( P_2 \)[/tex]. Rearrange Boyle's Law to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{P_1 \times V_1}{V_2} \][/tex]
Substitute the known values into the equation:
[tex]\[ P_2 = \frac{4.0 \text{ atm} \times 275 \text{ mL}}{175 \text{ mL}} \][/tex]
Calculate the numerator:
[tex]\[ 4.0 \text{ atm} \times 275 \text{ mL} = 1100 \text{ atm·mL} \][/tex]
Now, divide by the new volume:
[tex]\[ P_2 = \frac{1100 \text{ atm·mL}}{175 \text{ mL}} \][/tex]
Perform the division:
[tex]\[ P_2 = 6.285714 \text{ atm} \][/tex]
We can round this value to one decimal place to match the options provided in the question:
[tex]\[ P_2 \approx 6.3 \text{ atm} \][/tex]
So, the new pressure when the gas is compressed to 175 mL is:
c. 6.3 atm
[tex]\[ P_1 \times V_1 = P_2 \times V_2 \][/tex]
where:
- [tex]\( P_1 \)[/tex] is the initial pressure
- [tex]\( V_1 \)[/tex] is the initial volume
- [tex]\( P_2 \)[/tex] is the new (final) pressure
- [tex]\( V_2 \)[/tex] is the new (final) volume
Given the values:
- [tex]\( P_1 = 4.0 \text{ atm} \)[/tex]
- [tex]\( V_1 = 275 \text{ mL} \)[/tex]
- [tex]\( V_2 = 175 \text{ mL} \)[/tex]
We need to find [tex]\( P_2 \)[/tex]. Rearrange Boyle's Law to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{P_1 \times V_1}{V_2} \][/tex]
Substitute the known values into the equation:
[tex]\[ P_2 = \frac{4.0 \text{ atm} \times 275 \text{ mL}}{175 \text{ mL}} \][/tex]
Calculate the numerator:
[tex]\[ 4.0 \text{ atm} \times 275 \text{ mL} = 1100 \text{ atm·mL} \][/tex]
Now, divide by the new volume:
[tex]\[ P_2 = \frac{1100 \text{ atm·mL}}{175 \text{ mL}} \][/tex]
Perform the division:
[tex]\[ P_2 = 6.285714 \text{ atm} \][/tex]
We can round this value to one decimal place to match the options provided in the question:
[tex]\[ P_2 \approx 6.3 \text{ atm} \][/tex]
So, the new pressure when the gas is compressed to 175 mL is:
c. 6.3 atm
