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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 1.76 torr at 38 ∘c, how many grams of neon are in the sign? (the volume of a cylinder is πr2h.)

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jleaf
So there are a lot of steps and conversions to this problem that I'll try to explain.

First, this is an ideal gas law problem, so you use the equation PV=nRT, where P is pressure in kPa, V is volume in liters, n is the amount of gas in moles, R is the gas constant, and T is temperature in Kelvin.

Then, you realize everything they gave you is in different units, meaning everything has to be converted. 760 torr is equivalent to 101.3 kPa, which is used as a conversion factor. Celsius to Kelvin is just adding 273. The volume is found by finding the volume of the cylinder using pi×(r^2)×h (but you also have to make sure your units are the same since one of the measurements they give you is in meters and the other in centimeters). Then, since you need it in liters, convert that as well, knowing that 1000cm^3 is equivalent to 1 L.

Plug everything into the ideal gas law equation.

Then, since it asks for grams and not moles, you'll need to do one more conversion using the molar mass of neon (on the periodic table) to find the mass is about 0.005 grams.
Ver imagen jleaf
Cricetus

There are "0.00488 grams" of neon are in the sign. A further solution is below.

Given;

Diameter,

  • d = 2.5 cm

Length,

  • L = 5.5 m

Pressure,

  • P = 1.76 torr

Temperature,

  • T = 38°C

As we know,

→ [tex]V = \pi r^2 h[/tex]

      [tex]= 3.14\times \frac{(2.5\times 10^{-2})^2}{4}\times 5.5[/tex]

      [tex]= 27\times 10^{-4} \ m^3[/tex]

      [tex]= 2.7 \ liters[/tex]

Now,

→              [tex]PV = nRT[/tex]

→ [tex]0.0023\times 2.7=\frac{W}{20}\times 0.0821\times 310[/tex]

→        [tex]0.00621= W\times 1.2725[/tex]

→                [tex]W = \frac{0.00621}{1.2725}[/tex]

→                [tex]W = 0.00488 \ grams[/tex]

Thus the above answer is right.  

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