Question

A tightly sealed 5.0-l flask contains 781 mm hg of ar at 19 °c. the flask is heated until the pressure is doubled. what is the temperature of the gas?

Answer 1

The temperature of the gas is 311°C

Why is the gas temperature at constant volume when pressure is doubled?

Using Kay-Lussac's law of gases:

• P1/T1 = P2/T2

where P1 = 781 mmHg

P2 = 1562 mmHg

T1 = 19°C = 292 K

T2 = ?

making T2 subject of formula of gas laws

T2 = T1 × P2 / P1

T2 = 292 × 1562/781

T2 = 584 K or 311° C

Therefore, the temperature of the gas is 311°C

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