Question

If 67.5 mol of an ideal gas occupies 71.5 L at 47.00 °C, what is the pressure of the gas?

Answer 1

24.8 atm

Step-by-step explanation:

For this problem, first identify all of the given quantities, and the requested quantity:

Amount: 67.5 mol

Volume: 71.5 L

Temperature: 47.00 °C

Pressure: ??

Since we are given that the gas is an "ideal gas", we can apply the ideal gas law, which relates all of the above quantities:

`PV=nRT`, where

• P is the pressure,
• V is the Volume,
• n is the amount in moles,
• R is the ideal gas constant (the value of R is different depending on which units are used for Pressure and Volume), and
• T is the Temperature measured in Kelvin.

For Pressure in units of "atmospheres" and Volume in units of "Liters",
`R=0.0821(L \cdot atm)/(mol \cdot K)`

Recall that to convert Celsius to Kelvin, one must add 273.15 or use the equation
`T_C+273.15=T_K`

So
`T_K=(47.00 + 273.15)K=320.15K`, which is the temperature that we'll need for the Ideal Gas Law.

Since the Pressure is the unknown quantity, we can isolate P in the equation before substituting the known values:

`PV=nRT`

divide both sides by V...

`P=(nRT)/(V)`

Now, substitute and calculate:

`P=((67.5~mol)(0.0821(L \cdot atm)/(mol \cdot K))(320.15~K))/((71.5~L))`

`P=24.8138638111888~atm`

Accounting for significant digits,
`P=24.8~atm`

Answer 2

Ideal Gas Law: P = nRT/V
n = 67.5 mol
V = 71.5 L
T = 47°C + 273 K = 320 К
R = 0.0821

P = (67.5•0.0821•320) / 71.5 = 24.8 atm