Suppose a can contains a residual amount of gas at a pressure of 755 mm Hg and a temperature of 25 °C. What would the pressure be if the can were heated to 1155 °C?

P1/T1=P2/T2 temps in Kelvins

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To solve this problem, we can use the combined gas law equation:

(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)

P₁ = initial pressure (755 mm Hg)
V₁ = initial volume (assume constant since the problem does not mention any change in volume)
T₁ = initial temperature (25 °C + 273.15) in Kelvin
P₂ = final pressure (unknown)
V₂ = final volume (assume constant as mentioned above)
T₂ = final temperature (1155 °C + 273.15) in Kelvin

Since V₁ and V₂ are both assumed to be constant, we can remove them from the equation. Rearranging the equation, we have:

(P₁ / T₁) = (P₂ / T₂)

Now we can substitute the given values:

P₁ = 755 mm Hg
T₁ = 25 °C + 273.15 = 298.15 K
T₂ = 1155 °C + 273.15 = 1428.15 K

Substituting the values into the equation:

(755 mm Hg / 298.15 K) = (P₂ / 1428.15 K)

Now we can solve for P₂:

P₂ = (755 mm Hg / 298.15 K) * 1428.15 K

P₂ ≈ 3611.64 mm Hg

Therefore, if the can were heated to 1155 °C, the pressure inside the can would be approximately 3611.64 mm Hg.