A 2 liter vessel contains krypton under a pressure of 1.5 MPa. With a constant volume of gas, a certain amount

| February 7, 2021 | education

A 2 liter vessel contains krypton under a pressure of 1.5 MPa. With a constant volume of gas, a certain amount of heat was provided, this led to an increase in the gas temperature by 3 times. How much heat was provided to the gas?

Data: V – capacity of the vessel with krypton (V = 2 l; in the SI system V = 0.002 m3); P1 – initial pressure of krypton (P1 = 1.5 MPa = 1.5 * 10 ^ 6 Pa); Тк (final temperature) = 3Тн (initial temperature).

Const: i is the number of degrees of freedom for krypton (i = 3).

1) Initial temperature: P1 * V = m * R * Tn / MKr, from where we express: Tn = P1 * V * MKr / (m * R).

2) Temperature change: ΔТ = Тк – Тн = 3Тн – Тн = 2Тн = 2 * P1 * V * MKr / (m * R).

3) Heat transferred to krypton: Q = ΔU = 0.5 * i * m * R * ΔТ / MKr = 0.5 * i * m * R * (2 * P1 * V * MKr / (m * R)) / MKr = 0.5 * i * 2 * P1 * V = i * P1 * V = 3 * 1.5 * 10 ^ 6 * 0.002 = 9 * 10 ^ 3 J.

Answer: Krypton gas was provided with 9 kJ of heat.



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