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An engineer turns the temperature of a gas down, holding the volume constant at 1,250 L. Initially, the temperature of the gas was 683. C and had a pressure of 241. KPa. Determine the pressure of the gas when the temperature decreased to 159C

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

(P₁ x V₁)/T₁ = (P₂ x V₂)/T₂

Where P₁, V₁, and T₁ are the initial pressure, volume, and temperature of the gas, respectively. P₂ is the pressure we want to find, V₂ is the same as V₁, and T₂ is the final temperature of the gas.

Substituting the given values, we get:

(241 KPa x 1,250 L)/683 K = (P₂ x 1,250 L)/432 K

Simplifying and solving for P₂, we get:

P₂ = (241 KPa x 1,250 L x 432 K)/(683 K x 1,250 L) = 152.4 KPa

Therefore, the pressure of the gas when the temperature decreased to 159C is approximately 152.4 KPa.
User TheCyberXP
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