0.7249tm 14.96 12) Determine the number of moles contained in each of the following gas systems: a. 1.25 L of O₂ at 1.06 atm and 250. K PV=nRT n=PY (1006 atm x 1.25L) .0646 moles 0.0821 (Leatm/ kmal) x 250.K) P=1.06 atm R=0.0821 (L.atm/komel) V=125L T=250.K Ab. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C .030 moles 0.925 atm X 0.80 L 0.0821L.atm/mol = 0.06455542 (0646 Moles =0.030029646 x 300.15K) (030 mole V=0.80L P=0.925atm T 276=27+273.15 = 30045K PV=nRT n=pv R
0.7249tm 14.96 12) Determine the number of moles contained in each of the following gas systems: a. 1.25 L of O₂ at 1.06 atm and 250. K PV=nRT n=PY (1006 atm x 1.25L) .0646 moles 0.0821 (Leatm/ kmal) x 250.K) P=1.06 atm R=0.0821 (L.atm/komel) V=125L T=250.K Ab. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C .030 moles 0.925 atm X 0.80 L 0.0821L.atm/mol = 0.06455542 (0646 Moles =0.030029646 x 300.15K) (030 mole V=0.80L P=0.925atm T 276=27+273.15 = 30045K PV=nRT n=pv R
0.7249tm 14.96 12) Determine the number of moles contained in each of the following gas systems: a. 1.25 L of O₂ at 1.06 atm and 250. K PV=nRT n=PY (1006 atm x 1.25L) .0646 moles 0.0821 (Leatm/ kmal) x 250.K) P=1.06 atm R=0.0821 (L.atm/komel) V=125L T=250.K Ab. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C .030 moles 0.925 atm X 0.80 L 0.0821L.atm/mol = 0.06455542 (0646 Moles =0.030029646 x 300.15K) (030 mole V=0.80L P=0.925atm T 276=27+273.15 = 30045K PV=nRT n=pv R
Can you help me with number 12 a.) and number 12 b.)? Were this the correct answer that shows my work includes the significant figures? Also, can you help me with the Ideal gas law formula to plug in?
Transcribed Image Text:Final Pressure (P₂): 14 atm Initial temperature (Ta) = 300.ok Initial temperature (1₁) = 200.ok
Final volume (V₂)
(P) = 12 atm
Pavax Ta
та
Pa
8) If I initially have a gas at a pressure if 12 atm, a volume of 23 liters, and a temperature of 200.0 K, and
then I raise the pressure to 14 atm and increase the temperature to 300.0 K, what is the new volume of the
V₂ = P gas?
T2
12 atm x23L X300.0K
(200.0k X14atm)
= 29.57142857
30. L
pressure.
Initial volums
Use the combined gas law to solve the following problems:
Ti.Pa
30. L
Piv, Pava Pav₂ x ta
Ta
V2=
V₁. P₁. Ta
Pa Ti
TI
Ta
T₁
9) A gas takes up a volume of 17 liters, has a pressure of 2.3 atm, and a temperature of 299 K. If I raise the
temperature to 350 K and lower the pressure to 1.5 atm, what is the new yolume of the gas?
PIVI Pa Va
Final volume (V₂)
initial volume (V.) = 172
initial Pressure (P₁) = 2.3 atm
T₂
31 L
Final Pressure (P₂) = 1.5 atm
initial temperature (T) = 299k Final temperature (To) = 350k
TaxPixVi
Pa.va.ta
=
=
Pa deal Gas Law
TX.B₂
12=
-XT2 P₁V₁ - P₂V₂
Ta
P.V₁ Ta
=
2.3atm x 77LX 350 k)
11. Sa+m. 299K)
= 30.51282051
31L
Pa Ti
10) Calculate the pressure, in atmospheres, exerted by each of the following:
a. 250 L of gas containing 1.35 moles at 320 K. Pv=n·R·T, R=0.0821
Mole n=1.35 mote Temperture (1)=320K P = n.r.r
Volume v= 250L Pressure =
V
.14 atm
P=014 atm
0.1418688
b., 4.75 L of gas containing 0.86 moles at 300. K.
mole: 0.86 mole
Pressure:
14.9 L
volume (v): 4.75L
Temperature: 300k
4.5 atm
V=4.75L
T= 300k
/Leatm
p=4.459atm
n=0.86 mole R=0.0821 (K+mole)
P = 4.5atm
11) Calculate the volume, in liters, occupied by each of the following:
a. 2.00 moles of H₂ at 300. K and 1.25 atm.
PV=nRT
V=nRT
2.00 mole X0.0821 (2) × 300.0K
1.259tm
39.4 L
P=n·R·T
V
= 39. 408
39.4L
= 1.35 mol x 0.0821 Komole X 320K
250L
= 0.86 molx0.0821 (atm).
4.75X
(2·atm
0.425 mole (NH3) X0.0821 (Komple) X 310.15K_
b. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C
.030 moles 0.925 atm X 0.80 L
• atm
mole
b. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°C
PV=nRT₁ V = nRT n=0.425 mol T=37°C = 37+273.15-k = 310.15 K
atm
P
= 14.94738795
14.92
0.7249tm
12) Determine the number of moles contained in each of the following gas systems:
a. 1.25 L of O2 at 1.06 atm and 250. K
Pr=nRT = P (1.06 atm x 1.25L)
RT
0.0821 (Leatm/ kmal) X 250.K)
P=1.06 atm R=0.0821 (L.atm/K+mol)
.0646 moles
v1.25L
T=250.K
^=
Komor X 300k
=0.030029646
0.0821L.atm/mol x 300.15K) (030 mole
4.459326316
P= 1.25 atm
V=
T = 300.K
n = 2.001 mot
R=0.082L a+m/ Kime)
= 0.06455542
(.0646 Moles)
PV=nRT
V=0.80L
P=0.925atm
T=276=27+273.15= 30045K
(4.5atm)
= RY
n = PV
Definition Definition Law that is the combined form of Boyle's Law, Charles's Law, and Avogadro's Law. This law is obeyed by all ideal gas. Boyle's Law states that pressure is inversely proportional to volume. Charles's Law states that volume is in direct relation to temperature. Avogadro's Law shows that volume is in direct relation to the number of moles in the gas. The mathematical equation for the ideal gas law equation has been formulated by taking all the equations into account: PV=nRT Where P = pressure of the ideal gas V = volume of the ideal gas n = amount of ideal gas measured in moles R = universal gas constant and its value is 8.314 J.K-1mol-1 T = temperature
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