# At constant temperature, the pressure on 8.0 L of a gas increases from 1.0 atm to 4.0 atm. What will be the new volume of the gas?

Jul 8, 2017

Well, according to old Boyle's law $P \propto \frac{1}{V}$, and get ${V}_{2} = 2 \cdot L$.........

#### Explanation:

And if $P \propto \frac{1}{V}$ ($\text{temperature}$, and $\text{moles of gas}$ CONSTANT), then $P V = k$, where $k$ is some constant.....and so for the same amount of gas UNDER different conditions of $\text{pressure}$ and $\text{volume}$ but constant $\text{temperature}$, we may write............

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

.........when we solve for $k$.........

And we fill in the blanks, ${V}_{1} = 8.0 \cdot L$, ${P}_{1} = 1 \cdot a t m$, and ${P}_{2} = 4 \cdot a t m$...........

${V}_{2} = \frac{{P}_{1} {V}_{1}}{P} _ 2 = \frac{1 \cdot \cancel{a t m} \times 8.0 \cdot L}{4 \cdot \cancel{a t m}} = 2 \cdot L$.

Is it reasonable that volume should be so reduced under higher pressure?