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For a certain ideal gas, Cp = 36.9 J mol-1deg-1. What will be the final temperature (in oC) when 2.00 moles of the gas at 20oC and 15.0 atm are allowed to expand adiabatically and reversibly to 5.00 atm?

Answer :

Final answer:

An ideal gas cools when it expands adiabatically and reversibly. The final temperature of the gas can be calculated using the equation for adiabatic processes based on the initial temperature, final and initial pressures, and the heat capacity ratio. In this case, the final temperature is -80.3°C.

Explanation:

The question pertains to an adiabatic and reversible expansion of a certain ideal gas. 'Adiabatic' means that no heat is exchanged between the gas and its surroundings during the process and 'reversible' means the process can be reversed by an infinitely small change in a variable. The as the pressure decreases from 15.0 atm to 5.00 atm, the gas expands and cools.

The final temperature of this adiabatic process can be calculated using the equation T2 = T1 * (P2/P1)^((gamma-1)/gamma), where gamma (γ) is the heat capacity ratio, which is Cp/Cv. For an ideal gas, Cv = Cp - R = 36.9 J mol-1deg-1 - 8.314 J mol-1deg-1 = 28.586 J mol-1deg-1. So, gamma = Cp/Cv = 36.9/28.586 = 1.29. Substituting these values into the equation, we find T2 = 20°C * (5.00 atm / 15.0 atm)^((1.29 - 1) / 1.29). Converting this to Kelvin by adding 273.15 results in T2 = -80.3°C.

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