Can you help me with the number 12 question? Can you explain step by step formula for the ideal gas law? Can you put the formula to plug in the fraction to give the correct answer to the formula? 1Final volume (V₂)Cnitial temperature (TInitial Pressure (P₁) = 12 atmInitial volume (V₁) = 234Initial temperenture (T₁1) = 200KUse 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, andthen I raise the pressure to 14 atm and increase the temperature to 300.0 K, what is the new volume of thegas? V₂= P₁V₁T₂T₁. PaWP, VIPa Va230. L+.12 atm x 23LX3000K(14 atm x 200.ok)= 29.57142857(30.4V₂9) 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 thetemperature to 350 K and lower the pressure to 1.5 atm, what is the newP,V,Pa Vainitial volume (V₁) = 172initial pressure (P₁) = 2.3atminitial temperature (T) 299kТаP₁V₁12=(2.3atm) (17L) x (350K)(299)(1. Satm)P.VITAVan TR=0.0821110) Calculate the pressure, in atmospheres, exerted by each of the following:a. 250 L of gas containing 1.35 moles at 320 K. Pyntmole n = 1.35 mole Temperature(1): 320k P = nr Ivolume v= 250L Pressure (1):.14 atmFinal Pressure (Pa): 14atm31 L TIIdeal Gas Law PaxT₂.PNb. 4.75 L of gas containing 0.86 moles at 300, K.volume (v): 4.75L mble:0.86 malceTemperature 3ook Pressure:4.5 atmV=4.75L39.4 L"P=ARTP= 4.459 atmT= 300kn=0.86mok, R=0.082 Latmak (4.5 atm).-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 P=0.724a+m.b. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°Cn=0.425 mole14.9 LRTvolume of the gas?V=MRI2.00 mde X 0.0821 L.atm/ K-mal) x(3000K) P1.25atmb. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C.030 moles (0,925 atm x 0,80L)Final volume (Va)Final Pressure (P₂) = 1.5atmFinal temperature (1₂) = 350 K30.5128205 LVa=312n=P=0.14 atm= 0.86m 1X0.0821_molk X 300k4.75 LLoatmK.more,= 1.35 mol x 6.0821 Mork 320katom L250LPV=nRTPivitaT, Pa= 39.408(39.4८atmolT=370=37+273.15K-310.15K,0.425mole (NH) X 0.0821 Latm/K·mol x 310.15K0.724.atm12) 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 P=106am T=250.K(1.06 atm x 1.25 L)n = PVV=1.25L0.082.1 Latm mol K¹X 250.K=.0646 moles=0.06455542 (0646 moles)P= 1.25atmT-300.K1=2,00lmoleR=0.082(a+mknol'rXR=0.0821L. atm molkV=0.80LP=0,925 atmT=27°C= 27+2 73.15=300,150.0821 latm. mol¹ KX 300.15K = 0.030029646.030 MoleZ14.14015884(14.92)commanePV=ARTn = PVRTH

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Answered: 12) Determine the number of moles…

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Can you help me with the number 12 question? Can you explain step by step formula for the ideal gas law? Can you put the formula to plug in the fraction to give the correct answer to the formula?

1
Final volume (V₂)
Cnitial temperature (T
Initial Pressure (P₁) = 12 atm
Initial volume (V₁) = 234
Initial temperenture (T₁1) = 200K
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= P₁V₁T₂
T₁. Pa
WP, VI
Pa Va
2
30. L
+.
12 atm x 23LX3000K
(14 atm x 200.ok)
= 29.57142857
(30.4
V₂
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
P,V,
Pa Va
initial volume (V₁) = 172
initial pressure (P₁) = 2.3atm
initial temperature (T) 299k
Та
P₁V₁12=(2.3atm) (17L) x (350K)
(299)
(1. Satm)
P.VITA
Van T
R=0.08211
10) Calculate the pressure, in atmospheres, exerted by each of the following:
a. 250 L of gas containing 1.35 moles at 320 K. Pynt
mole n = 1.35 mole Temperature(1): 320k P = nr I
volume v= 250L Pressure (1):
.14 atm
Final Pressure (Pa): 14atm
31 L TI
Ideal Gas Law Pax
T₂.PN
b. 4.75 L of gas containing 0.86 moles at 300, K.
volume (v): 4.75L mble:0.86 malce
Temperature 3ook Pressure:
4.5 atm
V=4.75L
39.4 L
"P=ART
P= 4.459 atm
T= 300k
n=0.86mok, R=0.082 Latmak (4.5 atm).
-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 P=0.724a+m.
b. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°C
n=0.425 mole
14.9 L
RT
volume of the gas?
V=MRI
2.00 mde X 0.0821 L.atm/ K-mal) x(3000K) P
1.25atm
b. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C
.030 moles (0,925 atm x 0,80L)
Final volume (Va)
Final Pressure (P₂) = 1.5atm
Final temperature (1₂) = 350 K
30.5128205 L
Va=312
n=
P=0.14 atm
= 0.86m 1X0.0821_molk X 300k
4.75 L
Loatm
K.more,
= 1.35 mol x 6.0821 Mork 320k
atom L
250L
PV=nRT
Pivita
T, Pa
= 39.408
(39.4८
atmol
T=370=37+273.15K-310.15K,
0.425mole (NH) X 0.0821 Latm/K·mol x 310.15K
0.724.atm
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 P=106am T=250.K
(1.06 atm x 1.25 L)
n = PV
V=1.25L
0.082.1 Latm mol K¹X 250.K=
.0646 moles
=0.06455542 (0646 moles)
P= 1.25atm
T-300.K
1=2,00lmole
R=0.082(a+mknol'r
X
R=0.0821L. atm molk
V=0.80L
P=0,925 atm
T=27°C= 27+2 73.15=300,15
0.0821 latm. mol¹ KX 300.15K = 0.030029646.030 Mole
Z
14.14015884
(14.92)
commane
PV=ART
n = PV
RT
H

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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