If the pressure on a gas at -73°C is doubled but its volume is held constant, what will its final temperature be in degrees Celsius?

127°C

Explanation:

This excersise can be solved, with the Charles Gay Lussac law, where the pressure of the gas is modified according to absolute T°.

We convert our value to K → -73°C + 273 = 200 K

The moles are the same, and the volume is also the same:

P₁ / T₁ = P₂ / T₂

But the pressure is doubled so: P₁ / T₁ = 2P₁ / T₂

P₁ / 200K = 2P₁ / T₂

1 /2OOK = (2P₁ / T₂) / P₁

See how's P₁ term is cancelled.

200K⁻¹ = 2/ T₂

T₂ = 2 / 200K⁻¹  → 400K

We convert the T° to C → 400 K - 273 = 127°C

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