To solve this question, we'll use the combined gas law, which relates the pressure, volume, and temperature of a gas. The combined gas law is expressed as:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where \( P_1 \) and \( P_2 \) are the initial and final pressures, \( V_1 \) and \( V_2 \) are the initial and final volumes, and \( T_1 \) and \( T_2 \) are the initial and final temperatures (measured in Kelvin), respectively.
Let's identify what we have:
- \( V_1 = 10.0 \text{ L} \) (initial volume)
- \( P_1 \) (initial pressure, we just need to know it is increased by 3 times, the actual value doesn't matter since it will cancel out)
- \( T_1 \) (initial temperature, similarly as pressure, we only need the relative increase)
- \( P_2 = 3 \times P_1 \) (final pressure is triple the initial pressure)
- \( T_2 = 1.8 \times T_1 \) (final temperature is increased by 80% which means it is 180% of the initial temperature)
Now we want to find \( V_2 \), the final volume. We can rearrange the combined gas law to solve for \( V_2 \):
\[ V_2 = \frac{P_1 V_1}{T_1} \times \frac{T_2}{P_2} \]
Since we know that \( P_2 = 3 \times P_1 \) and \( T_2 = 1.8 \times T_1 \), we can plug them into the equation:
\[ V_2 = \frac{P_1 \times 10.0}{T_1} \times \frac{1.8 \times T_1}{3 \times P_1} \]
Simplifying the equation, we notice that \( P_1 \) and \( T_1 \) cancel out:
\[ V_2 = 10.0 \times \frac{1.8}{3} \]
Calculate the final volume \( V_2 \):
\[ V_2 = 10.0 \times 0.6 \]
\[ V_2 = 6.0 \text{ L} \]
Therefore, the new volume of the gas when the pressure is tripled and the temperature is increased by 80% will be 6.0 L.
