When the pressure of a gas is halved and its temperature is doubled, what happens to its volume?

1 Answer
Oct 9, 2017

The volume of the gas will increase by a factor of $2 \cdot 2 = 4$.

Explanation:

Your tool of choice here will be the combined gas law equation, which looks like this

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2}}}$

Here

• ${P}_{1}$, ${V}_{1}$, ${T}_{1}$ are the pressure, volume, and absolute temperature of the gas at an initial state
• ${P}_{2}$, ${V}_{2}$, ${T}_{2}$ are the pressure, volume, and absolute temperature of the gas at a final state

Now, notice that when the temperature is kept constant, increasing the pressure of the gas by a specific factor will cause its volume to decrease by the same factor $\to$ think Boyle's Law here.

Similarly, when the pressure is kept constant, increasing the temperature of the gas by a specific factor will cause its volume to increase by the same factor $\to$ think Charles' law here.

This tells you that increasing the temperature of the gas will cause its volume to increase. Similarly, decreasing its pressure will also cause its volume to increase.

So even without doing any calculations, you should be able to say that

${V}_{2} > {V}_{1}$

In other words, decreasing the pressure of the gas and increasing its temperature are changes that do not compete with each other in terms of their influence on the volume of the gas.

In your case, you have

${P}_{2} = {P}_{1} / 2 \to$ the pressure of the gas is halved

${T}_{2} = 2 \cdot {T}_{1} \to$ the temperature of the gas is doubled

Rearrange the combined gas law equation to solve for ${V}_{2}$

${V}_{2} = {P}_{1} / {P}_{2} \cdot {T}_{2} / {T}_{1} \cdot {V}_{1}$

Plug in your values to find

${V}_{2} = {\overbrace{\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{1}}}}}{\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{1}}}}}{2}}}}^{\textcolor{b l u e}{\text{influence of pressure")) * overbrace( (2 * color(red)(cancel(color(black)(T_1))))/color(red)(cancel(color(black)(T_1))))^(color(blue)("influence of temperature}}} \cdot {V}_{1}$

${V}_{2} = 2 \cdot 2 \cdot {V}_{1}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{V}_{2} = 4 \cdot {V}_{1}}}}$

As predicted, the volume of the gas increased as a result of the two changes. Notice that the volume increased by a factor that is equal to the product of the factor that corresponds to the decrease in pressure, i.e. $2$, and the factor that corresponds to the increase in temperature, i.e. $2$.