# You have a sample of neon gas at a certain pressure, volume, and temperature. You double the volume, double the number of moles of neon, and double the Kelvin temperature. How does the final pressure (Pf) compare to the original pressure (Po) ?

Mar 10, 2015

The final pressure will double as well.

If you use the ideal gas law equation, $P V = n R T$, to express the initial and the final states of your gas sample, you can write something like this

${P}_{1} {V}_{1} = {n}_{1} \cdot R \cdot {T}_{1}$ $\to$ the initial state; (1)
${P}_{2} {V}_{2} = {n}_{2} \cdot R \cdot {T}_{2}$ $\to$ the final state;

You also know that the volume doubles, which means that ${V}_{2} = 2 \cdot {V}_{1}$, the number of moles doubles, ${n}_{2} = 2 \cdot {n}_{1}$, and the temperature doubles, ${T}_{2} = 2 \cdot {T}_{1}$. Use these equalities into the equation that describes the final state

${P}_{2} \cdot 2 \cdot {V}_{1} = 2 \cdot {n}_{1} \cdot R \cdot 2 \cdot {T}_{1}$ (2)

Divide equation (1) by equation (2) and you'll get

${P}_{1} / \left(2 \cdot {P}_{2}\right) = \frac{{n}_{1} \cdot R \cdot {T}_{1}}{2 \cdot {n}_{1} \cdot R \cdot 2 \cdot {T}_{1}}$

${P}_{1} / \left(2 \cdot {P}_{2}\right) = \frac{1}{4} \implies 2 \cdot {P}_{2} = 4 \cdot {P}_{1} \implies {P}_{2} = 2 \cdot {P}_{1}$

Therefore, the pressure doubles as well. Finally, the answer makes sense because the temperature and the number of moles will each double the pressure, but the fact that the volume doubles as well will cancel one of these effects out.

If the volume would have stayed the same, the pressure would have increased 4 times - 2 times because the temperature doubled and 2 times more because the number of moles doubled.