Question

A light bulb contains argon gas at a temperature of 295 k and at a pressure of 75 kilopascals. the light bulb is switched on, and after 30 minutes its temperature is 418 k. what is a correct numerical setup for calculating the pressure of the gas inside the light bulb at 418 k?

Answer 1

Normally for these types of questions (2 different situations), you can use:
P*V/T = P*V/T
As long as you don't confuse the different values up
Since V is constant, you can say that P/T (before heating up) = P/T (after heating up)
So 75000/295 = P/418

Answer 2

Answer: The pressure of the gas inside the light bulb at 418 K is 106.27 kilo pascal.

Explanation:

$P_1=75$ kilo Pascals

$T_1=295 K$

$P_2=?$

$T_2=418 K$

According Gay Lussac's law:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$P_2=\frac{P_1\times T_2}{T_2}=\frac{75 \text{kilo pascal} \times 418 K}{295 K}=106.27$ kilo pascal

The pressure of the gas inside the light bulb at 418 K is 106.27 kilo pascal.