A 10.2g 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 5.25L volume and the temperature raised to 62 ∘C, what is the new gas pressure?

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.k1.mol1
(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 = 1.31 = 0.3 atm
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