Respuesta :
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
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