1. A 160 L kiln is used for vutrifying ceramics. It is currently operating at 11

Question

1. A 160 L kiln is used for vutrifying ceramics. It is currently operating at 1145 degrees Celsius and the pressure is 1.100 atm. How many miles of air molecules are within the confines of the kiln
Secure I https//session.mast Chap 11 Introduction to the Ideal Gas Law y.com/myct/itemView?assignmentProblemiD 87 ·prev ± r- luction to the Ideal Gas Law Part A Learning Goal A balloon is floating around outside your window. The temperature outside is 29 C, and the air pressure is 0.700 atm Your neighbor, who released the baloon, tells you that he filed it with 3.80 moles of gas. What is the volume of gas inside this balloon? To apply the ideal gas law to problems involving temperature, pressure, volume, and moles of a gas Express your answer to three significant figures and include the appropriate units The four properties of gases (pressure P volume V temperature T and moles of gas n) are related by a single expression known as the ideal gas law Hints PV = nRT The variable R is known as the universal gas constant and has the value R = 00821 L . atm/(mole·K) The unts of R dictate the units for all other quantities, so when using this value of R use units of atmospheres for pressure, liters for volume, and kelvins for temperature KlValue Units Submit My Answers Give Up Part B A 200 L gas cylinder is iled with 820 moles of gas The tank is stored at 7 "C. What is the pressure in the tank? press your answer to three significant figures and include the appropriate units. Hints PkValue Units Submit My Answers Give Up

Explanation / Answer

PV= nRT

P = Pressure in atm

V= Volume in Liter

n = no of moles

R = 0.0821 L atm K-1 Mol-1

T = Temperature in Kelvin

Question a

P = 0.7 atm , V =?

n = 3.8 moles T = 273 + 29 = 302 K

V = nRT / P

V = 3.8 moles x 0.0821 L atm K-1 Mol-1 x 302 K / 0.7 atm

V = 134.597 Liter

Question b

V= 20 Liter n = 8.2 Moles

T = 273 +7 = 280 K

P = nRT / V

P = 8.2 moles x 0.0821 L atm K-1 Mol-1 x 280 K / 20 Liter

P = 9.425 atm

.Question c

V = 160 Liter T = 273 +1143 = 1416 K

P = 1.1 atm , n = ?

n = PV / RT

n = 1.1 atm x 160 Liter / 0.0821 L atm K-1 Mol-1 x 1416 K = 1.51 Moles

Question d

V = 18 Liter n = 4.6 Moles

P = 3.4 atm T=?

T = PV / nR

T = 3.4 x 18 / 0.0821 x 4.6 = 162.05 K or 111.01 Deg Celcius

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