Question

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

Answer 1

Answer:

              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

Answer 2

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