Final Pressure (P₂): 14 atm Initial temperature (Ta) = 300.ok Initial temperature (1₁) = 200.ok Final volume (V₂) Initial Pressure. Initial volume (V.) = ressure 23L (Pi) = 12 atm Pavax Ta та Pa Use the combined gas law to solve the following problems: 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 gas? |V₂=1₁V₁+2= 200.0k X14 atm) 12 atm X234 X 3000K = 29.57142857 30. L Ti-Pa 30. L
Final Pressure (P₂): 14 atm Initial temperature (Ta) = 300.ok Initial temperature (1₁) = 200.ok Final volume (V₂) Initial Pressure. Initial volume (V.) = ressure 23L (Pi) = 12 atm Pavax Ta та Pa Use the combined gas law to solve the following problems: 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 gas? |V₂=1₁V₁+2= 200.0k X14 atm) 12 atm X234 X 3000K = 29.57142857 30. L Ti-Pa 30. L
Final Pressure (P₂): 14 atm Initial temperature (Ta) = 300.ok Initial temperature (1₁) = 200.ok Final volume (V₂) Initial Pressure. Initial volume (V.) = ressure 23L (Pi) = 12 atm Pavax Ta та Pa Use the combined gas law to solve the following problems: 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 gas? |V₂=1₁V₁+2= 200.0k X14 atm) 12 atm X234 X 3000K = 29.57142857 30. L Ti-Pa 30. L
Can you help me with number 8? Was this the correct answer that shows my work includes the significant figures? Also, can you help me with the combined 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 Any of various laws that describe the ways in which volume, temperature, pressure, and other conditions correlate when matter is in a gaseous state. At a constant temperature, the pressure of a particular amount of gas is inversely proportional with its volume (Boyle's Law) In a closed system with constant pressure, the volume of an ideal gas is in direct relation with its temperature (Charles's Law) At a constant volume, the pressure of a gas is in direct relation to its temperature (Gay-Lussac's Law) If the volume of all gases are equal and under the a similar temperature and pressure, then they contain an equal number of molecules (Avogadro's Law) The state of a particular amount of gas can be determined by its pressure, volume and temperature (Ideal Gas law)
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