Question

1) A sample of He has a volume of 0.180 L, a pressure 0.800 atm and a temperature of 302 K. At what temperature (K) will the He have a volume of 90.0 mL and a pressure of 3.20 atm. (Hint: Combined Gas Law)

Reasoning:

Solution:

2) A cylinder contains 5.0 L of oxygen gas at 20.0 OC and 0.850 ATM. How many grams of oxygen are in the cylinder? (Hint: Ideal Gas Law)

Reasoning:

Solution:

Please show all work! Please and thank you!

Answer 1

(1) We know that PV = nRT

Where

T = Temperature ;P = pressure ; n = No . of moles ;R = gas constant ; V= Volume of the gas

As the gas remains the same , n will constant

So PV / T = constant

Thereby PV/T = P'V'/T'

Where

P = initial pressure = 0.8 atm

V = initial volume = 0.180 L

T = initial temperature = 302 K

P'= final pressure = 3.20 atm

V' = final volume = 90.0 mL = 0.09 L

T' = final temperature = ?

Plug the values we get T' = (P'V'T) / (PV) = 604 K

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(2)

We know that ideal gas equation is PV = nRT

Where

T = Temperature = 20 oC = 20 + 273 = 293 K

P = pressure = 0.850 atm

n = No . of moles = ?

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = 5.0 L

PLug the values we get n = (PV) / (RT)

                                     = 0.177 moles

So mass of oxygen , m = number of moles x molar mass

                                   = 0.177 mol x 32 g/mol

                                   = 5.65 g

Therefore the mass of oxygen in the cylinder is 5.65 g