Question

Ideal gas Laws / Gas Stoichiometry:

a) A light bulb contains .56 g of Argon gas at 1.34 atm at 26 deg C. The light bulb is then turned on and, after a period of time, the temperature rises to 179 deg C. Calculate the final pressure inside the light bulb.

b) How many liters of oxygen at STP are needed to burn 3.0 L of octane?

Answer 1

A)

we know that

PV = nRT

the amount of argon (n) and volume of the bulb remains constant

so

V and n are constant

so

we get

P1/T1 = P2/T2

given

P1 = 1.34

T1 = 26 + 273 = 299 K

T2 = 179 + 273 = 452 K

so

using those values

we get

1.34 / 299 = P2 / 452

P2 = 2.0257

so

final pressure inside the bulb is 2.0257 atm

b)

the balanced reaction is

C8H18 + 12.5 02 ---> 8 CO2 + 9 H20

we can see that

moles of oxygen = 12.5 x moles of octane

so

moles of oxygen / moles of ocatane = 12.5

now

PV = nRT

both octane and Oxygen are at STP

so

P , T , R are same for both

so

V/T = constant

(n / V ) of ocatne = (n/V) of Oxygen

volume of oxygen / volume of octane = moles of oxygen / moles of octane

so

volume of oxygen / volume of octane = 12.5

volume of oxygen / 3 = 12.5

volume of oxygen = 37.5

so

37.5 L of oxygen is required to react with 3 L of octane