Question

A certain quantity of ideal gas takes up 56J of heat in the process $AB$ and 360J in the process $AC$. What is the number of degrees of freedom of the gas.

Answer 1

5

Answer 2

For process AB (isobaric): $Q_{AB} = 56$ J $W_{AB} = p_0(3V_0 - V_0) = 2p_0V_0$ $\Delta U_{AB} = \frac{f}{2}(p_0(3V_0) - p_0V_0) = fp_0V_0$ Using the First Law of Thermodynamics ($Q = \Delta U + W$): $56 = fp_0V_0 + 2p_0V_0 = (f+2)p_0V_0$ (1)

For process AC (linear, $P \propto V$): $Q_{AC} = 360$ J $W_{AC} = \int_{V_0}^{4V_0} \frac{p_0}{V_0}V dV = \frac{15}{2}p_0V_0$ $\Delta U_{AC} = \frac{f}{2}(4p_0(4V_0) - p_0V_0) = \frac{15f}{2}p_0V_0$ Using the First Law of Thermodynamics: $360 = \frac{15f}{2}p_0V_0 + \frac{15}{2}p_0V_0 = \frac{15}{2}(f+1)p_0V_0$ (2)

Dividing (2) by (1): $\frac{360}{56} = \frac{\frac{15}{2}(f+1)p_0V_0}{(f+2)p_0V_0}$ $\frac{45}{7} = \frac{15}{2} \frac{f+1}{f+2}$ $\frac{6}{7} = \frac{f+1}{f+2}$ $6(f+2) = 7(f+1)$ $6f + 12 = 7f + 7$ $f = 5$