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Edit View History Bookmarks People Window Help MasteringChemistry: Bonus 2 (Chap 5 ecure https://session.masteringchemistry.com/myct/itemView?assignmentProblemID-8 us 2 (Chap 5-8 Item 13 Part A Calculate the volume, in liters, of 5.80×10-3 mol of a gas at 31°C and 0.500 atm . Express your answer with the appropriate units. vValue Units Submit My Answers Give Up Part B Calculate the pressure, in atmospheres, of 1.74x10-2 mol CH4 (g) in a 0.295 L flask at 38C. Express your answer with the appropriate units. P- Value Units Submit My Answers Give Up

Explanation / Answer

part A

We know that PV = nRT

V = nRT/P

n = 5.80 X 10-3 mole,

T = 310C = 31+273.15 = 304.15K,

P= 0.500 atm,

R = 0.08205 L atm mol-1 K-1 ( R = gas constant)

V = ?

Substitute these value in above equation.

V = 5.80 X 10-3 X 0.08205 X 304.15/0.500 = 0.2895 L

volume of gas = 0.2895 Liter

Part B

Ideal gas equation

PV = nRT             where, P = atm pressure= ?

V = volume in Liter = 0.295 L

n = number of mole = 1.74 X 10-2 mol

R = 0.08205L atm mol-1 K-1 =Proportionality constant = gas constant,

T = Temperature in K = 380C = 273.15+ 38 = 311.15 K

We can write ideal gas equation

P = nRT/V

Substitute value in above equation

P = 1.74 X10-2 X0.08205X 311.15 / 0.295 = 1.50 atm

pressure of methane = 1.50 atm

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