12=2.3 atm) (174) (350K) (299) Satm) Va=312 ^P2 √₂= Ideal Gas Law PXTI 10) Calculate the pressure, in atmospheres, exerted by each of the following: 1 a. 250 L of gas containing 1.35 moles at 320 K. Pv=nrt R=0.0821 Fatm mok! mole n = 1.35 mole Temperature (1): 320k P = nr I volume v= 250L Pressure (1): .14 atm P=₂ b. 4.75 L of gas containing 0.86 moles at 300, K. volume (v): 4.75L mole:0.86 malc Temperature 3ook Pressure: 4.5 atm V=4.75L T= 300k 0=12186 mak 2- An = nrt 30.5128205 L P= 4.459 atm mix 14.5 9+m = 1.35 mol x 0.0821 atm mol KX320k 250L P=0.14a+m) = 0.86m 1x0.082 | Latm mol k X 300 k 4.75L
12=2.3 atm) (174) (350K) (299) Satm) Va=312 ^P2 √₂= Ideal Gas Law PXTI 10) Calculate the pressure, in atmospheres, exerted by each of the following: 1 a. 250 L of gas containing 1.35 moles at 320 K. Pv=nrt R=0.0821 Fatm mok! mole n = 1.35 mole Temperature (1): 320k P = nr I volume v= 250L Pressure (1): .14 atm P=₂ b. 4.75 L of gas containing 0.86 moles at 300, K. volume (v): 4.75L mole:0.86 malc Temperature 3ook Pressure: 4.5 atm V=4.75L T= 300k 0=12186 mak 2- An = nrt 30.5128205 L P= 4.459 atm mix 14.5 9+m = 1.35 mol x 0.0821 atm mol KX320k 250L P=0.14a+m) = 0.86m 1x0.082 | Latm mol k X 300 k 4.75L
12=2.3 atm) (174) (350K) (299) Satm) Va=312 ^P2 √₂= Ideal Gas Law PXTI 10) Calculate the pressure, in atmospheres, exerted by each of the following: 1 a. 250 L of gas containing 1.35 moles at 320 K. Pv=nrt R=0.0821 Fatm mok! mole n = 1.35 mole Temperature (1): 320k P = nr I volume v= 250L Pressure (1): .14 atm P=₂ b. 4.75 L of gas containing 0.86 moles at 300, K. volume (v): 4.75L mole:0.86 malc Temperature 3ook Pressure: 4.5 atm V=4.75L T= 300k 0=12186 mak 2- An = nrt 30.5128205 L P= 4.459 atm mix 14.5 9+m = 1.35 mol x 0.0821 atm mol KX320k 250L P=0.14a+m) = 0.86m 1x0.082 | Latm mol k X 300 k 4.75L
Can you help me with the number 10 question? Can you explain step by step including the formula (Ideal Gas Law)? I need to plug in the fraction to give the correct answer.
Transcribed Image Text:Initial Pressure (P₁) = 12 atm
Initial volume (V₁) = 234
Initial temperenture (T₁) = 2500.0
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
V.P, MI
gas?
V₂= Pivita
тра
atm x 23LX3000K
Pava
Fa
30. L
(14 atm x 200.0k).
T₁
29/57142857
30.4
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
va-Pivita
T. Pa
temperature to 350 K and lower the pressure to 1.5 atm, what is the new volume of the gas?
volume = 172
initial pressure (P) = 2.3atm Final Pressure (Pa) = 1.5 atm
initial temperature (T) 299K
Final temperature (12) = 350 K
30.5128205 L
Final Pressure (Pa): 14atm
Final volume (Va) - 300K
Initial temperature (1₂)
P.V₁ - Pa Va
Та
31 L
√₂=RK x JB
PiVita - 3 atm) (174) X(350K)
Рахті
(2991)
(1,Satm)
Pa v₂=
Ideal Gas Law
39.4 L
10) Calculate the pressure, in atmospheres, exerted by each of the following:
a. 250 L of gas containing 1.35 moles at 320 K. Pynrt R=0.0821 fatm mo'k!
mole n = 1.35 mole Temperature (1): 320k P = nr I
volume v= 250L Pressure (1):
.14 atm
P=nrt
b. 4.75 L of gas containing 0.86 moles at 300, K.
volume (v): 4.75L mole:0.86 Molec
Temperature 300k Pressure:
P= 4.459 atm
4.5 atm
V=4.75L
n=0.86mok, R=0.082 Latmma1 / 14.5 atm)
T= 300k
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.
4
2.00mde X 0.0821 Latm mol K x 300k
1.25atm
b. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°C
PV=nRT, V = nRÌ_n=0.425 mole
atm
12=
Va= 312
P=0.14 atm
= 0.86m 1X0.082 | Latm mol k X300K
4.75 La
= 1.35 mol x 0.082 1 atm mol K²X 320k
250L
RT
PV=nRT
n=
V=MRI
= 39.408
39.42
T-376-37+273 k= 310k
14.9 L 0.425 mole (NH3) X 0.0821 Latm mol¹k'x (273+37)k.
a. 1.25 L of O₂ at 1.06 atm and 250. K PV=ART
(1.06 atm x 1,25 L)
n = PV
0.0821 Latm mot klx 250 K=1
.0646 moles
-0.06455542.0646 moles
b. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C V0.80L
030 moles, 25 atm x 0.80L)
P=125atm
に
T= 300.K
6.724 atm
12) Determine the number of moles contained in each of the following gas systems:
P=106 atm T=250.k
V=1.252
1=2,00lmole
R=0.082(a+mk mol's
7
14.940 15884
(14.92)
R=0.08212. atm molk
P=0,925 atm
T=27°C= 21+ 273.15=300.15
0.08211atm. mol" K'x 300.15K = 0.030029646030 Mole
PV = nRT
n = PV
RT
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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