# The pressure of 4.2 L of nitrogen gas in a flexible container is decreased to one-half its original pressure, and its absolute temperature is increased to double the original temperature. What is its volume now?

Dec 27, 2015

$\text{17 L}$

#### Explanation:

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

$\textcolor{b l u e}{\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2} \text{ }$, where

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

Now, even without doing any calculations, you can examine the above equation and predict what you expect the volume of the gas to be at the final state.

Notice that pressure and volume have an inverse relationship when temperature is kept constant - this is known as Boyle's Law.

This tells you that a decrease in pressure will cause an increase in volume.

On the other hand, temperature and volume have a direct relationship when pressure is kept constant - this is known as Charles' Law.

This tells you that an increase in temperature will cause an increase in volume.

In your case, these two effects will actually combine. Increasing the temperature of the gas while decreasing its pressure will result in an increase in volume, regardless of the actual changes in temperature and pressure.

Mathematically, you can prove this by rearranging the combined gas law equation to solve for ${V}_{2}$

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

You know that

• ${P}_{2} = {P}_{1} / 2 \to$ the pressure is halved
• ${T}_{2} = 2 \cdot {T}_{1} \to$ the temperature is doubled

This means that you have

${V}_{2} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{1}}}}}{\frac{1}{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{1}}}}} \cdot \frac{2 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{T}_{1}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{T}_{1}}}}} \cdot {V}_{1}$

${V}_{2} = 2 \cdot 2 \cdot {V}_{1} = 4 \cdot {V}_{1}$

Indeed, both effects combine to cause an increase in volume.

The volume of the gas will thus be

${V}_{2} = 4 \cdot \text{4.2 L" = "16.8 L}$

Rounded to two sig figs, the answer will be

${V}_{2} = \textcolor{g r e e n}{\text{17 L}}$