7. How many moles of argon are there in 20.0 L, at 25 degrees Celsius and 96.8 kPa?

The moles of argon that are there in 20.0 L, at 25 c and 96.8 KPa is = 0.7814 moles

calculation

by use of the ideal gas equation

that is PV=nRT where;

P (pressure) = 96.8 kpa

V (volume) = 20.0 L

n (moles) = ?

R (gas constant) = 8.314 kpa/k/mol

T (temperature) = 25 + 273 = 298 K

make n the formula of the subject

n = PV/RT

n is therefore = [ (96.8 KPa x 20.0 l) / (8.314 kpa/k/mol x 298 k) ]

answer = 0.781 moles

0.8133 mol

Solution:

Data Given:

Moles = n = ?

Temperature = T = 25 °C + 273.15 = 298.15 K

Pressure = P = 96.8 kPa = 0.955 atm

Volume = V = 20.0 L

Formula Used:

Let's assume that the Argon gas is acting as an Ideal gas, then according to Ideal Gas Equation,

P V = n R T

where; R = Universal Gas Constant = 0.082057 atm. L. mol⁻¹. K⁻¹

Solving Equation for n,

n = P V / R T

Putting Values,

n = (0.955 atm * 20.0 L) : (0.082057 atm. L. mol⁻¹. K⁻¹ * 298.15 K)

n = 0.8133 mol