Question

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.

Answer 1

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: $n=7.16*10^-^4moles$

V = 75.0 mL = 0.0750 L

P = 116.8 kPa

We know that, 101.325 kPa = 1 atm

So, $116.8kPa(\frac{1aym}{101.325kPa})$

= 1.15 atm

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

T = ?

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

$1.15(0.0750)=7.16*10^-^4*0.0821(T)$

0.08625 = 0.00005878(T)

$T=\frac{0.08625}{0.00005878}$

T = 1467 K

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

Answer 2

Answer : The temperature of the light bulb was $1.47 \times 10^3K$

Explanation :

Using ideal gas equation :

$PV=nRT$

where,

P = pressure of gas = 116.8 kPa

V = volume of gas = 75.0 mL = 0.075 L      (1 L = 1000 mL)

T = temperature of gas = ?

n = number of moles of gas= $7.16\times 10^{-4}mole$

R = gas constant = $8.314\text{L kPa }mol^{-1}K^{-1}$

Now put all the given values in the ideal gas equation, we get:

$(116.8kPa)\times (0.075L)=(7.16\times 10^{-4}mole)\times (8.314\text{L kPa }mol^{-1}K^{-1})\times T$

$T=1471.6K\approx 1.47 \times 10^3K$

Thus, the temperature of the light bulb was $1.47 \times 10^3K$