Hand lighters contain butane gas. If a hand lighter contains butane gas at 2.50 atm of pressure at 293°K, then what would be the new pressure of the lighter if it were left in a
hot car at 393°K? If the lighter was designed to hold only 3.00 atm of pressure, would
it explode? Assume the volume does not change, until it possibly explodes!

3.35 atm

Since P₂ > 3.00 atm, the lighter would explode.

Explanation:

Step 1: Given data

Initial pressure of butane gas (P₁): 2.50 atmInitial temperature of butane gas (T₁): 293 KFinal pressure of butane gas (P₂): ?Final temperature of butane gas (T₂): 393 K

Step 2: Calculate the final pressure of butane gas

If we assume ideal behavior, we can calculate the final pressure of butane gas using Gay Lussac's law.

P₁/T₁ = P₂/T₂

P₂ = P₁ × T₂/T₁

P₂ = 2.50 atm × 393 K/293 K = 3.35 atm

Since P₂ > 3.00 atm, the lighter would explode.

Dmight be the correct answer

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