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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?

Verified Answer :

amyseal

Normally for these types of questions (2 different situations), you can use:
PV/T = PV/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: 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.

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