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Assuming the volume is constant, if a gas has an initial pressure of 84.0 kPa at 30.0°C, what would the new pressure be at 240.0°C?

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The new pressure at 240.0 * C would be 142 kPa
User Oskar Krawczyk
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Answer:

Step-by-step explanation:

Gay-Lussac's Law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant.

P₁=84kPa

P₂=?kPa

The temperatures have first been converted to Kelvin.

T1=30°C

Converting to kelvin; 30+273= 303K

T2=240°C

Converting to kelvin; 240+273 = 513K

Step 1: Use Gay-Lussac's Law to solve for the unknown pressure (P₂).


(P_1 )/T_1 =P_2/T_2

Step 2: Solve.

First, rearrange the equation algebraically to solve for P₂.


P_2= (P_1 x T_2)/T_1

Now substitute the known quantities into the equation and solve.


P_2=(84 x 513)/303

P₂ = 142kPa

User Asterius
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