what temperature will 215 mL of a gas at 20°C and 1 atm pressure attain when it is subject to 1.5 atm of pressure?

Given:

Volume of gas = 215 ml

Initial temperature T1 = 20C = 20 + 273 = 293 K

Initial pressure P1 = 1 atm

Final pressure P2 = 1.5 atm

To determine:

The final temperature T2

Explanation:

Based on Gay-Lussac's Law:

P α T

Therefore we have the relation:

P1/T1 = P2/T2

T2 = P2T1/P1 = 1.5 * 293/1 = 439.5 K

Converting from Kelvin to degrees C we have:

T2 = 439.5 - 273 = 166.5 C

Ans: The final temperature is 439.5 K or 166.5 C