Chemistry - Gas Law Problems. Thanks in advance!!!?

Question

These are a few problems I have. Please just tell me what law to use and what number is what. Thanks!

1. A stainless steel tank has a volume of 25 L. When the tank held 15.4 moles of helium, the pressure of the gas was 755 torr. The tank developed a small leak and lost 4.2 moles of gas. If the temperature remained unchanged, what was the pressure in the tank after the loss?

2. What is the volume of H2 gas requred to produce 7.53 liters of ammonia gas (NH3) at STP?
N2 + 3H2 --> 2NH3

3. The pessure of the air in an automobile tire was 32 lbs/in^2 when the tire was cool (20 degrees Celsius). After being driven for a while, the tire pressure was 38 lbs/in^2 . If the volume of the tire stays constant and no leaks develop, what is temperature of the air in the tire after driving?

4. 138 grams of Carbon Disulfide (CS2) vapors at STP were allowed to expand into an empty 50.0 L tank. After being moved to a location in the sun, the pressure was 1.14 atm. What is the temperature?

Answer 1

Lancenigo di Villorba (TV), Italy

TROUBLE 1)
Since you refer to Ideal Gas as Helium it is, I apply the "Ideal Gas law"

(p1 * V1) / (n1 * T1) = (p2 * V2) / (n2 * T2)

The experiment interests a "Throttle's Flow" of a gas from its vessel. The Temperature and the Volume remain unchanged, hence I may state

p1 / n1 = p2 / n2

retrieving

p2 = n2 * p1 / n1 = (15.4 - 4.2 ) * 755 / 15.4 = 549 torr

TROUBLE 2)
The experiment interests "Ammonia Synthesis"

N2(g) + 3 H2(g) <---> 2 NH3(g)

starting from its chemical elements. When I refer to Standard Temperature/Pressure Conditions (e.g. 0°C and 1 atm), I understand that 7.53 STP-liter of Ammonia require

3 * 7.53 / 2 = 11.29 STP-liter

11.29 liter of Hydrogen.

TROUBLE 3)
Since you refer to Ideal Gas as Air may behave, I apply the "Ideal Gas law"

(p1 * V1) / (n1 * T1) = (p2 * V2) / (n2 * T2)

The experiment interests an "Isochoric Charles law" of a gas in its vessel. The Volume remain unchanged, hence I may state

p1 / T1 = p2 / T2

retrieving

T2 = T1 * p2 / p1 = 293.16 * 38 / 32 = 345 K = 75 °C

TROUBLE 4)
Since you refer to Ideal Gas as Carbon Sulphide may behave, I apply the "Ideal Gas law"

p1 * V1 = n1 * R * T1

in order to detemine the volume occupied by CS2's 138 g in the Standard Temperature/Pressure Conditions (273.16K and 1 atm), that is

V1 = n1 * R * T1 / p1 = (138 / 76) * 0.0821 * 273.16 / 1 =
= 40.7 STP-liter

The experiment interests an "Polytropic Phenomenon" as Gas-Expansion in a greater vessel, hence I rearrange the "Ideal Gas law" in this manner

(p1 * V1) / T1 = (p2 * V2) / T2

retrieving

T2 = (T1 * p2 * V2) / (p1 * V1) =
= (273.16 * 1.14 * 50) / (1 * 40.7) = 382 K = 108 °C

I hope this helps you.

Answer 2

PV=nRT can be used to solve all of these, it is just a matter of figuring out the correct relationships.

PV=nRT rearranges to R=PV/nT. Thus, with these types of before and after problems, you know the P1V1/n1T1 = P2V2/n2T2. Know you don't need to memorize Boyles Law, Charles Law, or Avogadros Law. You can solve any gas law problem with the relationship given above.

1. Temperature and Volume are constant before and after the leak. You can use the relationship P1/n1 = P2/n2

2. This is a limiting reagent problem. Use PV=nRT to determine the moles of NH3 gas. Then use mole ratio (3/2) to determine the amount of H2 needed.

3. Your constants are volume and moles. This is an application of Boyles Law. You can use the relationship P1/T1 = P2/T2 to solve this problem.

4. Use PV=nRT again. You have P, V, n (after an easy calculation), and R as knowns. This is a plug and chug problem.

Answer 3

1) you remember pV = nRT so p =nRT/V anp is proportionnal to n . In your case the value of n dropped 15.4-4.2 =11.2moles
so the pressure is p/755 =11.2/15.4
p = 755*11.2/15.4 =549 torr

2) you need 3 moles of H2 for 2 moles of NH3. the moles are proportional to volume so you need 7.5 *3/2 =11.25 L of H2

3) as in the first exercice the pressure is proportionnal to T
so T(final)/T(initial)= 38/32 Tinitial ) 273+20=293K
Tfinal = 293*38/32=348K

4)pV = nRT T = pV/nR

p= 1.14*101325=115510 Pa n = 138/(12+2*32) = 138/76 mole V = 0.05m^3 r=8.31

T=383K