Ideal Gas Law XT₁ 1.5 atm x 2.99 k R=0.082 nrT 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 mole n = 1.35 mle Temperature(t): 30k P = nr I volume v= 250L Pressure (1): T .14 atm b. 4.75 L of gas containing 0.86 moles at 300, K. -olume (v): 4.75L mole:0.86 mole Temperature: Book Pressure: 4.5 atm v₂ = 3 P=4.459 atm Va-E = 1.35 m. P=0.14 atm nrt = 0.86 m 1x0.0° 4.
Ideal and Real Gases
Ideal gases obey conditions of the general gas laws under all states of pressure and temperature. Ideal gases are also named perfect gases. The attributes of ideal gases are as follows,
Gas Laws
Gas laws describe the ways in which volume, temperature, pressure, and other conditions correlate when matter is in a gaseous state. The very first observations about the physical properties of gases was made by Robert Boyle in 1662. Later discoveries were made by Charles, Gay-Lussac, Avogadro, and others. Eventually, these observations were combined to produce the ideal gas law.
Gaseous State
It is well known that matter exists in different forms in our surroundings. There are five known states of matter, such as solids, gases, liquids, plasma and Bose-Einstein condensate. The last two are known newly in the recent days. Thus, the detailed forms of matter studied are solids, gases and liquids. The best example of a substance that is present in different states is water. It is solid ice, gaseous vapor or steam and liquid water depending on the temperature and pressure conditions. This is due to the difference in the intermolecular forces and distances. The occurrence of three different phases is due to the difference in the two major forces, the force which tends to tightly hold molecules i.e., forces of attraction and the disruptive forces obtained from the thermal energy of molecules.
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Initial Pressure (P₁) = 12 atm
Initial volume (V₁) = 23L
Final Pressure (Pa)
Final volume V
Initial temperature
Use the combined
gas
- 300k
law to solve
the following problems: temperature (T₁) = 200K
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 = Pivita
P₁V₁
V₂-12am x23LX300.0k
Pa Va
Ta
14 atm x 200.0k/
тра
29.57142857
BOL
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 volume of the gas?
Final volume (Va)
·P,V,
Pa Va
30. L
31 L
TI
v₂= P₁Vita= 2,3atmx 17LX350k
Ideal Gas Law PXT
1.5 atm x 2,99 K
v₂ =
V=312]
10) Calculate the pressure, in atmospheres, exerted by each of the following:
a. 250 L of gas containing 1.35 moles at 320 K. Pnrt R=0.0821 1 atm mo [¹
molen = 1.35 mole Temperature (1): 320k P = nr I
volume v= 250L Pressure (1):
.14 atm
b. 4.75 L of gas containing 0.86 moles at 300, K.
volume(v): 4.75L mole:0.86 mole
Temperature: 300k Pressure:
4.5 atm
initial volume (V₁) = 172
initial pressure (P₁) = 2.3 atm
initial temperature (T) 299K
39.4 L
14.9 L
P= nrt
V
P= 4.459 atm
(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.
b. 0.425 moles of ammonia gas (NH3) at 0.724 atm and 37°C
.0646 moles
Final Pressure (P₂) = 1.5atm
Final temperature (1₂)= 350 k
30.51282051
b. 0.80 L of ammonia gas (NH3) at 0.925 atm and 27°C
.030 moles
= 1.35 mol x 0.082 1 atm mol kx320k
250L
P=0.14 atm
= 0.86m 1X0.0821 Latm mol¹k X 300k
4.75 L
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
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