One mole ideal monoatomic gas is heated according to path AB and AC . If temperature of state B and state C are equal.

Question

TestHub · Accepted Answer

Process AC = polytropic process.

The molar heat capacity $c_{m} = c_{v} + R /2 = 2 R$ .

For process AB, which is isobaric, $c_{m} = c_{p} = 5 R /2$ .

The ratio of heat transferred is:

$\frac{q _{A C}}{q _{A B}} = \frac{\int _{T_{A}}^{T_{C}} n C _{m} \cdot d T}{\int _{T_{A}}^{T_{B}} n \cdot C _{p, m} \cdot d T} = \frac{2 R}{\frac{5}{2} R} = 0.8$

Chemistry - Thermodynamics - 1 Question with Solution | TestHub

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