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?

Answer: 1470 K

Explanation: 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:

V = 75.0 mL = 0.0750 L

P = 116.8 kPa

We know that 101.325 kPa = 1 atm

So,

= 1.15 atm

R is universal gas constant and its value is.

T = ?

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

0.08625 = 0.00005878 (T)

T = 1467 K

So, the temperature of the light bulb would be 1467 K, but since we have to round the total to equal 3 significant figures, the answer is 1470 K

n = number of mole of sample of helium gas = 7.16 x 10⁻⁴ moles

T = temperature of the gas = temperature of the bulb?

V = Volume of helium gas = 75 mL = 7.5 x 10⁻⁵ m³

P = pressure of the helium gas = 116.8 kPa = 116.8 x 10³ Pa (1 k = 1000)

Using the equation

P V = n R T

inserting the values

(116.8 x 10³) (7.5 x 10⁻⁵) = (7.16 x 10⁻⁴) (8.314) T

T = 1471.6 C