What gas law should I use to solve the following problems?

Question

A 6.53g of solid CO2 is put in an empty sealed 7.00 L container at a temperature of 310 K. When all the solid CO2 becomes gas, what will be the pressure in the container?

A cylinder with a movable piston contains 5.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 4.10 L to 6.70 L? (The temperature was held constant.)

Answer 1

Part I

Given :-

Mass of solid CO₂ = 6.53 gm

Volume of container, V = 7.00 L

Temperature, T = 310 K

Now we have to calculate the number of moles of CO₂

No. of moles = mass / molar mass

[we do know the value of molar mass of CO₂ which is 44.01 g/mol]

No. of moles, n = 6.53 g / 44.01 g/mol

n = 0.1484 mol

Using the ideal gas equation, PV = nRT

[we do know the value of the R constant which is 0.08206 L.atm K^-1 mol^-1]

P = nRT / V

P = 0.1484 mol x 0.08206 L.atm K^-1 mol^-1 x 310 K / 7.00 L

P = 0.5393 atm

Hence, the pressure in the container is 0.5393 atm.

Part II

Given :-

Mass of Helium(m1) = 5.0 gm

Temperature remains same ie (T) is constant

Pressure remains same ie. (P) is constant

Volume (V1) = 4.10 L

Volume (V2) = 6.70 L

According to Ideal gas law,

  PV = n RT                              (where R is a constant)

 

  PV/nT = constant 

P1V1 / n1T1 = P2V2 / n2T2                                   (P is constant & T is constant) 

V1 / n1 = V2 / n2 

since n = mass / mw 

V1 / (mass 1 / mw) = V2 / (mass 2 / mw) 

and since molecular mass = constant 

V1 / mass 1 = V2 / mass 2 

mass 2 = V2 x mass 1 / V1

= 6.70L x 5.00 g / 4.10 L

mass 2 = 8.17 g 

= mass 2 - mass 1

= 8.17 g - 5.00 g

= 3.17 g 

ANSWER : 3.17 g of He was added.

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Answer 2

Look to see what information is given, and what is being asked for.

a).

Given:

mass of CO2, so we can calculate moles = n = 6.53 g CO2 x 1 mol / 44 g = 0.148 mols

Volume = 7.00 L

Temperature = 310K

Asked for:

Pressure = P

USE THE IDEAL GAS LAW, PV = nRT

Solving the problem:

P = nRT/V = (0.148 mol)(0.0821 Latm/Kmol)(310K) / 7.00 L

P = 0.540 atm

b)

Given:

Initial moles of He = 5.00 g x 1 mol / 4 g = 1.25 mols = n₁

Initial volume = 4.10 L = V₁

Final moles of He = ?

Final volume = 6.70 L = V₂

USE V₁/n₁ = V₂/n₂ (Avogadro's law)

4.10 L / 1.25 mols = 6.70 L / x mols

x = 2.04 mols

2.043 mol He x 4 g / mol = 8.17 g final mass

Amount of He added = 8.17 g - 5.00 g = 3.17 g He added