Question

This is on gas laws!

Show all the work in solving the problems please!!
1. Find the pressure in mmHg of 3.2 moles of methane in a 250 mL container at 45C.
2. Find the volume in Liters of 4.25 moles of oxygen gas at 27C and a pressure of 5.05 atm.
3. Find the mass of O2 in a 4.5 liter container at 302kPa and 15C
4. Find the Temperature in a container at a pressure of 810 mmHg ,a avloie of 2.0 liter with 0.56 moles of gas.
5. Find the moles of nitrogen gas in a 55Liter container at 3.22 atm and 17C.

Answer 1

See Explanation

Step-by-step explanation:

These problems are based upon the Ideal Gas Law, PV = nRT. Variables are gas phase variables and include ...

Pressure (P)

Volume (V)

Mass (moles (n))

R = Gas Constant = 0.08206L·atm/mol·K

Temperature (T)

Ideal Gas Law => PV = nRT

Note => when using the gas constant, R, convert all dimensional units to match those of L·atm/mol·K.

Problem 1 => Given ...

P = nRT/V

V = 250 ml = 0.250 L

n = 3.2 moles

R = 0.08206L·atm/mol·K

T = 45°C = (45 + 273)K = 318K

P = nRT/V = (3.2mol)(0.08206L·atm/mol·K)(318K)/(0.250L) = 334atm = 334atm(760mmHg/atm) = 253,853mmHg ≅2.5 x 10⁵mmHg (2 sig. figs.)

Problem 2 = Given ...

P = 5.05atm

V = nRT/P

n = 4.25 moles

R = 0.08206L·atm/mol·K

T = 27°C = (27 + 273)K = 300K

V = (4.25mol)(0.08206L·atm/mol·K)(300K)/(5.05atm) = 20.7 Liters O₂(g)

Problem 3 = Given ...

P = 302 KPa = (302KPa)(0.01atm/KPa) = 3.02 atm

V = 4.5 Liters

n = PV/RT

R = 0.08206L·atm/mol·K

T = 15°C = (15 + 273)K = 288K

n = (3.02atm)(4.5L)/(0.08206L·atm/mol·K)(288k) = 0.58 mole O₂ = 0.58 mole x 32 g/mole = 18.4 grams O₂

Problem 4 = Given ...

P = 810mmHg = 810mmHg(1atm/760mmHg) = 1.07atm

V = 2.0 Liters

n = 0.56 mole N₂(g)

R = 0.08206L·atm/mol·K

T = PV/nR = (1.07atm)(2.0L)/(0.56mol)(0.08206L·atm/mol·K) = 46.6K = (46.6 - 273)°C = -226°C

Problem 5 = Given ...

P = 3.2 atm

V = 55 Liters

n = PV/RT

R = 0.08206L·atm/mol·K

T = 17°C = (17 + 273)K = 290K

n = (3.2atm)(55L)/(0.08206L·atm/mol·K)(290K) = 7.4 moles N₂(g)