A gas with a volume of 1.17 L at a pressure of 264.5 kPa is allowed to expand to a volume of 11.25 L. What is the pressure in the container if the temperature remains constant?

Give your answer in kPa to the tenths place. Do not put units in the answer space.



Answer :

To solve the problem of finding the final pressure of a gas when it expands, we can use Boyle's Law, which states that the pressure of a gas times its volume is a constant value as long as the temperature remains constant. Mathematically, Boyle's Law 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 final pressure
- [tex]\( V_2 \)[/tex] is the final volume

Given the values:
- [tex]\( V_1 = 1.17 \)[/tex] liters
- [tex]\( P_1 = 264.5 \)[/tex] kPa
- [tex]\( V_2 = 11.25 \)[/tex] liters

We need to find [tex]\( P_2 \)[/tex].

Step-by-step solution:

1. Start with Boyle's Law formula:
[tex]\[ P_1 \times V_1 = P_2 \times V_2 \][/tex]

2. Substitute the known values into the equation:
[tex]\[ 264.5 \text{ kPa} \times 1.17 \text{ L} = P_2 \times 11.25 \text{ L} \][/tex]

3. Solve for [tex]\( P_2 \)[/tex] by isolating it on one side of the equation:
[tex]\[ P_2 = \frac{264.5 \text{ kPa} \times 1.17 \text{ L}}{11.25 \text{ L}} \][/tex]

4. Perform the multiplication of the numerator first:
[tex]\[ 264.5 \times 1.17 = 309.465 \][/tex]

5. Now divide by the final volume to get [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{309.465}{11.25} \][/tex]

6. Complete the division:
[tex]\[ P_2 = 27.5 \text{ kPa} \][/tex]

Therefore, the pressure in the container after the gas has expanded to 11.25 liters, keeping the temperature constant, is 27.5 kPa.