Answer:
0.001mol
Explanation:
First let us determine the new volume
P1 = 225kpa
V1 = 2.25L
T1 = 25°C = 25 + 273 = 298k
P2 = 185 kPa
T2 = 10°C = 10 +273 = 283k
V2 =?
P1V1/T1 = P2V2/T2
225x2.25/298 = 185xV2/283
298x185xV2 = 225x2.25x283
V2 = (225x2.25x283) / (298x185)
V2 = 2.6L
Now let us find the number of mole at each state
State 1:
V = 2.25L
P = 225kpa = 225000Pa we must covert this to atm
101325Pa = 1atm
225000Pa = 225000/101325 = 2.22atm
T = 25°C = 25 + 273 = 298k
R = 0.082atm.L/mol/K
n =?
PV = nRT
n = PV/RT
n = 2.25x2.22/0.082x298
n = 0.204mol
State 2:
V = 2.6L
P = 185 kPa = 185000Pa we must covert this to atm
101325Pa = 1atm
185000Pa = 185000/101325 = 1.83atm
T = 10°C = 10 +273 = 283k
R = 0.082atm.L/mol /k
n =?
PV = nRT
n = PV/RT
n = 1.83x2.6/0.082x283
n = 0.205mol
Change in number of mole
= 0.205 — 0.204 = 0.001mol
The moles of gas that were lost are:
0.001 mol
It is given by: PV=nRT
Given:
P₁ = 225 kPa
V₁ = 2.25L
T₁ = 25°C = 25 + 273 = 298k
P₂ = 185 kPa
T₂ = 10°C = 10 +273 = 283k
To find:
V₂ =?
P₁* V₁ / T₁ = P₂* V₂ / T₂
225* 2.25/298 = 185* V₂ /283
298* 185* V₂ = 225* 2.25* 283
V₂= (225* 2.25* 283) / (298* 185)
V₂ = 2.6L
Calculation for number of moles for each state:
For State 1:
V = 2.25L
P = 225kpa = 225000Pa
101325 Pa = 1atm
225000Pa = 225000/101325 = 2.22atm
T = 25°C = 25 + 273 = 298k
R = 0.082atm.L/mol/K
n =?
PV = nRT
n = PV/RT
n = 2.25* 2.22 /0.082 *298
n = 0.204mol
For State 2:
V = 2.6L
P = 185 kPa = 185000 Pa
101325 Pa = 1atm
185000 Pa = 185000/101325 = 1.83atm
T = 10°C = 10 +273 = 283k
R = 0.082atm.L/mol /k
n =?
PV = nRT
n = PV/RT
n = 1.83 * 2.6/ 0.082* 283
n = 0.205mol
Change in number of mole: 0.205 — 0.204 = 0.001mol
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