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A 75.0-milliliter lightbulb is filled with neon. There are 7.16 × 10-4 moles of gas in it, and the absolute pressure is 116.8 kilopascals after the bulb has been on for an hour. How hot did the bulb get?

The temperature of the lightbulb was
K.

Respuesta :

Answer:- 1467 K

Solution:- It asks to calculate the kelvin temperature of the light bulb. Looking at the given info, it is based on ideal gas law equation, PV=nRT.

Given: [tex]n=7.16*10^-^4moles[/tex]

V = 75.0 mL = 0.0750 L

P = 116.8 kPa

We know that, 101.325 kPa = 1 atm

So, [tex]116.8kPa(\frac{1aym}{101.325kPa})[/tex]

= 1.15 atm

R is universal gas constant and it's value is [tex]0.0821\frac{atm.L}{mol.K}[/tex] .

T = ?

Let's plug in the values in the equation and solve it for T.

[tex]1.15(0.0750)=7.16*10^-^4*0.0821(T)[/tex]

0.08625 = 0.00005878(T)

[tex]T=\frac{0.08625}{0.00005878}[/tex]

T = 1467 K

So, the temperature of the light bulb would be 1467 K.

Eduard22sly

The temperature of the lightbulb is 1467 K

Data obtained from the question

  • Volume (V) = 75 mL = 75 / 1000 = 0.075 L
  • Number of mole (n) =  7.16×10¯⁴ mole
  • Pressure (P) = 116.8 KPa = 116.8 / 101.325 = 1.15 atm
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Temperature (T) =?

How to determine the temperature

The temperature of the bulb can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

Divide both side by nR

T = PV / nR

T = (1.15 × 0.075) / (7.16×10¯⁴ × 0.0821)

T = 1467 K

Therefore, the temperature of the bulb is 1467 K.

Learn more about ideal gas equation:

https://brainly.com/question/4147359

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