# 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?

##### 1 Answer

#### 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.

#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#

Here you have

*universal gas constant*, usually given as

**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

#PV = nRT implies T = (PV)/(nR)#

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 = "1297 K"#

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

#color(purple)(|bar(ul(color(white)(a/a)color(black)(t[""^@"C"] = T["K"] - 273.15)color(white)(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**.