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Heat is added to two identical samples of a monatomic ideal gas. In the first sample, the heat is added while the volume of the gas is kept constant, and the heat causes the temperature to rise by 80 K. In the second sample, an identical amount of heat is added while the pressure (but not the volume) of the gas is kept constant. By how much does the temperature of this sample increase

Answer :

Answer:

The temperature of the sample increase by 48 Kelvin

Explanation:

The sample is identical.

Hence the heat at constant pressure is equal to the heat at the constant Volume

Q1 = Q2

Q 1 = heat at constant pressure

Q2 = heat at the constant Volume

Substituting the given values, we get -

[tex]\frac{3}{2} nRT_1 = \frac{5}{2} nRT_2\\3 * 80 = 5 * T_2\\T_2 = 48[/tex]

The temperature of the sample increase by 48 Kelvin

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