Question
A neon sign is made of glass tubing whose inside diameter is 2.5 cm
and whose length is 5.5 m
.If the sign contains neon at a pressure of 1.74 torr
at 38 ∘C
, how many grams of neon are in the sign? (The volume of a cylinder is πr2h
.)
and whose length is 5.5 m
.If the sign contains neon at a pressure of 1.74 torr
at 38 ∘C
, how many grams of neon are in the sign? (The volume of a cylinder is πr2h
.)
Answers
GPT 3.5
To find the number of grams of neon in the sign, we first need to calculate the volume of neon gas using the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, we need to convert the pressure to atm:
1.74 torr = 1.74/760 atm = 0.00228 atm
Next, we calculate the volume of the neon gas in the sign:
r = 2.5 cm / 2 = 1.25 cm = 0.0125 m
h = 5.5 m
V = π(0.0125 m)^2 * 5.5 m
V ≈ 0.0027 m^3
Now we can rearrange the ideal gas law to solve for the number of moles:
n = PV / RT
n = (0.00228 atm)(0.0027 m^3) / (0.0821 L atm / K mol)(311 K)
n ≈ 0.00023 mol
Finally, we can calculate the mass of neon in the sign using the molar mass of neon (20.18 g/mol):
mass = n * molar mass
mass = 0.00023 mol * 20.18 g/mol
mass ≈ 0.0047 g
Therefore, there are approximately 0.0047 grams of neon in the neon sign.
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, we need to convert the pressure to atm:
1.74 torr = 1.74/760 atm = 0.00228 atm
Next, we calculate the volume of the neon gas in the sign:
r = 2.5 cm / 2 = 1.25 cm = 0.0125 m
h = 5.5 m
V = π(0.0125 m)^2 * 5.5 m
V ≈ 0.0027 m^3
Now we can rearrange the ideal gas law to solve for the number of moles:
n = PV / RT
n = (0.00228 atm)(0.0027 m^3) / (0.0821 L atm / K mol)(311 K)
n ≈ 0.00023 mol
Finally, we can calculate the mass of neon in the sign using the molar mass of neon (20.18 g/mol):
mass = n * molar mass
mass = 0.00023 mol * 20.18 g/mol
mass ≈ 0.0047 g
Therefore, there are approximately 0.0047 grams of neon in the neon sign.
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