Question

The pressure in an engine cylinder increases from 7.6 atm to 13 atm as the gas is compressed from 0.50 L to 0.10 L. What is the expected cylinder temperature if the gas was initially 15°C?

Answer 1

T k = 15 + 273 = 288 K

4.6 / 13 => 0.353 atm

0.50 / 0.10 => 5 L

(15 + 273) K x (13 atm / 7.6 atm) x (0.50 L / 0.10 L) 

= 2463.15 K

hope this helps!

Answer 2

Answer:

98.53K or 174.47°C

Explanation:

Initial Pressure, P1 = 7.6atm

Final Pressure, P2 = 13atm

Initial Volume, V1 = 0.50L

Final Volume, V2 = 0.10L

Initial Temperature, T1 = 15 +273 = 288K (Converting to kelvin temperature)

Final Temperature, T2 = ?

The equation relating all these parameters id the ideal gas equation.

This is given as;

(P1V1) / T1 = (P2V2) / T2

Upon solving for T2, we have;

T2 = P2V2T1 / PIV1

Inserting the values,

T2 = (13 * 0.10 * 288) / (7.6 * 0.5)

T2 = 374.4 / 3.8

T2 = 98.53K or 174.47°C