Question

A 10.2-g sample of a gas has a volume of 5.25 L at 27 ∘C and 764 mm Hg .

Part A

If 3.0 g of the same gas is added to this constant 5.25-L volume and the temperature raised to 62 ∘C, what is the new gas pressure?

Answer 1

from ideal gas equation

PV = nRT

PV = (W/M)*RT

M = molarmass of gas = ? g/mol

w = wt of gas in the sample = 10.2 g

P = 764 mmhg = 764/760 = 1 atm

V = 5.25 L

T = 273+27 = 300 k

R = 0.0821 l.atm.k-1.mol-1

(1*5.25) = (10.2/M)*0.0821*300

M = molarmass of gas = 47.85 g/mol

after addition of 3.0 g of gas,

Pressure(P) = nRT/V

             = ((10.2+3)/47.85)*0.0821*300/5.25

             = 1.3 atm

raise in pressure = 1.3-1 = 0.3 atm