PLEASE EXPLAIN EACH. 1a.) 14.0 g of oxygen gas occupies 14.5 L in an expandable

Question

PLEASE EXPLAIN EACH.

1a.) 14.0 g of oxygen gas occupies 14.5 L in an expandable container at a particular temperature and pressure. What volume will the the gas occupy if 42.0 g of oxygen are added to container. Assume the pressure and temperature are held constant.

L?

1b.) 28.5 g of nitrogen gas exerts a pressure of 1.50 atm in a rigid container at a particular temperature. What pressure will the the gas exert if 57.0 g of nitrogen are added to container. Assume the volume and temperature are held constant.

atm?

1c.) What is the volume of a mixture of 26.0 g of N2 and 29.0 g of O2 at 48.0°C and 720. mm Hg? Assume the gases do not react with each other.

Explanation / Answer

Question 1a

PV = nRT

V = nRT / P

V2/V1 = (n2(O2)RT / P) / (n1(O2)RT / P) =>

V2/V1 = n2(O2) / n1(O2)

V2/14.5L = n2(O2) / n1(O2)

n1(O2) = 14 g / 32g/mol = 0.44 mol

n2(O2) = (14 + 42) g / 32g/mol = 1.75 mol

V2 = (14.5 L × 1.75 mol) / 0.44 mol = 57.67 L

Question 1b

PV = nRT

P = nRT / V

P2/P1 = (n2(N2)RT / V) / (n1(N2)RT / V) =>

P2/P1 = n2(N2) / n1(N2)

P2/1.5 atm = n2(N2) / n1(N2)

n1(N2) = 28.5 g / 28g/mol = 1.0 mol

n2(N2) = (28.5 + 57) g / 28 g/mol = 3.0 mol

P2 = (1.5 atm × 3.0 mol) / 1.0 mol = 4.5 atm

Question 1c

n(N2) = 26.0 g / 28g/mol = 0.93 mol

n(O2) = 29.0 g / 32g/mol = 0.91 mol

n total = 0.91 + 0.93 = 1.84 mol

PV = nRT

V = nRT / P = (1.84 mol × 62.37 L×mmHg/mol×K × (48 + 273)K) / 720 mmHg  

V = 51.16 L

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