Question
A volume of gas at 1.1atm was measured at 0.326 L. What will the volume be when changed to 1.9atm?
Answers
GPT-4o mini
To find the new volume of the gas when the pressure changes, we can use Boyle's law, which states that the pressure of a gas is inversely proportional to its volume when temperature and the amount of gas are held constant. This relationship can be expressed with the formula:
\[ P_1 \times V_1 = P_2 \times V_2 \]
Where:
- \( P_1 \) is the initial pressure (1.1 atm),
- \( V_1 \) is the initial volume (0.326 L),
- \( P_2 \) is the final pressure (1.9 atm),
- \( V_2 \) is the final volume (what we are trying to find).
We can rearrange this formula to solve for \( V_2 \):
\[ V_2 = \frac{P_1 \times V_1}{P_2} \]
Now plug in the values:
\[ V_2 = \frac{1.1 \, \text{atm} \times 0.326 \, \text{L}}{1.9 \, \text{atm}} \]
Calculating that:
\[ V_2 = \frac{0.3596 \, \text{atm} \cdot \text{L}}{1.9 \, \text{atm}} \]
\[ V_2 = 0.1897 \, \text{L} \]
Therefore, the new volume of the gas at 1.9 atm will be approximately **0.190 L**.
\[ P_1 \times V_1 = P_2 \times V_2 \]
Where:
- \( P_1 \) is the initial pressure (1.1 atm),
- \( V_1 \) is the initial volume (0.326 L),
- \( P_2 \) is the final pressure (1.9 atm),
- \( V_2 \) is the final volume (what we are trying to find).
We can rearrange this formula to solve for \( V_2 \):
\[ V_2 = \frac{P_1 \times V_1}{P_2} \]
Now plug in the values:
\[ V_2 = \frac{1.1 \, \text{atm} \times 0.326 \, \text{L}}{1.9 \, \text{atm}} \]
Calculating that:
\[ V_2 = \frac{0.3596 \, \text{atm} \cdot \text{L}}{1.9 \, \text{atm}} \]
\[ V_2 = 0.1897 \, \text{L} \]
Therefore, the new volume of the gas at 1.9 atm will be approximately **0.190 L**.