# If 2.35 moles of an ideal gas has a pressure of 3.63 atm, and a volume of 68.93 L, what is the temperature of the sample in degrees Celsius?

Jun 8, 2016

${1020}^{\circ} \text{C}$

#### Explanation:

This looks like a job for the ideal gas law, which establishes a relationship between the volume, pressure, number of moles, and temperature of a gas.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Here you have

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

Notice that the ideal gas law equation uses absolute temperature, which is another term used for temperature expressed in Kelvin.

In your case, the problem wants you to find the temperature in degrees Celsius, so right from the start you know that you must convert the temperature from Kelvin to degrees Celsius.

So, rearrange the ideal gas law equation and solve for $T$

$P V = n R T \implies T = \frac{P V}{n R}$

The units given to you for pressure and volume match those used in the expression of the universal gas constant, so plug them into the equation to find

T = (3.63 color(red)(cancel(color(black)("atm"))) * 68.93color(red)(cancel(color(black)("L"))))/(2.35color(red)(cancel(color(black)("moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * "K"))

$T = \text{1297 K}$

Now you're ready to convert this to degrees Celsius by using

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{t \left[\text{^@"C"] = T["K}\right] - 273.15} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in your value to find

t = "1297 K" - 273.15 = color(green)(|bar(ul(color(white)(a/a)color(black)(1020^@"C")color(white)(a/a)|)))

The answer is rounded to three sig figs.