# 500.0 L of gas are prepared at 0.921 atm pressure and 200.0°C. The gas is placed into a tank under high pressure. When the tank cools to 20.0°C, the pressure is 30.0 atm. What is the volume of the gas under these conditions?

Jul 4, 2018

$\text{V" = 24.8 color(white)(l) "L}$

#### Explanation:

The ideal gas law suggests the relationship

$P \cdot V = n \cdot R \cdot T$

that relates

• $P$, the pressure of the gas
• $V$, the volume the gas occupies
• $n$, the number of moles of gas in the closed container
• $T$, the temperature of the gas in $\textcolor{p u r p \le}{\text{degrees Kelvin}}$, and
• $R$ the ideal gas constant.

Let ${n}_{1}$, ${P}_{1}$, ${V}_{1}$, and ${T}_{1}$ resembles the respective properties of the gas before the change and ${n}_{2}$, ${P}_{2}$, ${V}_{2}$, and ${T}_{2}$ properties after the change.

• ${n}_{1} = \frac{R \cdot {T}_{1}}{{P}_{1} \cdot {V}_{1}}$

• ${n}_{2} = \frac{R \cdot {T}_{2}}{{P}_{2} \cdot {V}_{2}}$

The amount of gas present in the container, $n$, stays the same during this process. ${n}_{1} = {n}_{2}$ and therefore

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

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

Multiplying both sides of the equation by $\frac{{P}_{1} \cdot {V}_{2}}{{T}_{1}}$ gives the relationship between the volume of the gas before and after the change:

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

Therefore

${V}_{2} = {V}_{1} \cdot \left(\frac{{T}_{2}}{{T}_{1}} \cdot \frac{{P}_{1}}{{P}_{2}}\right)$
color(white)(V_2) = 500.0 color(white)(l) "L" xx ((200 + 273.15) color(white)(l) color(red)(cancel(color(purple)(K)))) / ((20 + 273.15) color(white)(l) color(red)(cancel(color(purple)(K)))) * (0.921 color(white)(l) color(red)(cancel(color(black)("atm")))) / (30.0 color(white)(l) color(red)(cancel(color(black)("atm"))))
$\textcolor{w h i t e}{{V}_{2}} = 24.8 \textcolor{w h i t e}{\text{L}}$
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