Question

A 75.9 g sample of O₂ gas at 0.0 ^oC and 380 Torr is compressed and heated until the volume is 2.94 L and the temperature is 27 ^oC. What is the final pressure in Torr?

Olmsted, Chemistry, 3e, Canadian Edition Intro to Chemical Properties of Matter etice Assignment Gradebook ORION Downloadable eTextbook ament Question 2 A 75.9 sample of o, gas at 0.0 °C and 380 Torr is comi d and heated until the volume is 2.94 L and the temperature is 27 °C. What is the final pressure in Tom Tam The number of significant digter is set to 3; the tolerance By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor Question Attempts: 0 of 3 used SAVE FOR LATES Earn Maximum Points aveable only if you answer this question correctly in two attempts or less

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Answer 1

mass of oxygen m=75.9 g

molar mass of oxygen M = sum of mass of oxygen atom =2 *16 g =32 g

atomic weight of oxygen atom = 16 g

moles =mass taken/molar mass

no of moles of O₂ (n)= m/M=75.9/32=2.372

universal gas constant R=0.08206 atm L K^-1 mol^-1

initial temperature T₁= 0^oC =273 +0= 273K

initial pressure P₁ =380 torr=380*0.00131579 atm = 0.5 atm [1torr=0.00131579atm]

initial volume =V₁

using ideal gas equation

P₁V₁ =nRT₁

V₁₌ =nRT_1/P₁

v₁ =2.372 mol* 0.08206 atm L K^-1 mol^-1 *273 K/0.5 atm

V₁ =106.277 L

final temperature T₁= 27^oC =273 +27= 300K

final pressure = P₂

final volume V₂₌ 2.94L

using combined gas equation

$P_{1}V_{1}/T_{1}=P_{2}V_{2}/T_{2}$

$(P_{1}V_{1}/T_{1})T_{2}/V_{2}=P_{2}$

(0.5 atm*106.277 L/273K)(300K/2.94L)=P₂

0.195 atm*300/2.94=P₂

58.5/2.94 atm=P₂

19.90 atm= P₂

P₂₌ 19.90 atm= 15124 torr [1torr=0.00131579atm]