Answer:
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
