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A cylinder of gas has a pressure of 440 kPa at 25*c. At what temperature in *c will it reach 650 kPa

User Vedanshu
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1 Answer

4 votes

Answer:

The answer is 167.29 °C

Step-by-step explanation:

According to Gay-Lussac's Law: The temperature and Pressure of a given gas are directly proportional to each other at given volume and amount.

The formula is as:

P₁ / T₁ = P₂ / T₂ -----(1)

In given case:

T₁ = 25 °C = 298.15 K

T₂ = Unknown

P₁ = 440 kPa

P₂ = 650 kPa

Solving equation 1 for T₂,

T₂ = P₂ × T₁ / P₁

Putting Values,

T₂ = 650 kPa × 298.15 K / 440 kPa

T₂ = 440.448 K

Or,

T₂ = 167.29 °C ∴ °C = K - 273.15

User Sunil Kumar Jha
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