Question

A certain quantity of an ideal gas takes up $56J$ of heat in the process $AB$ and $360J$ in the process $AC$, as shown in the graph below. The adiabatic constant for the gas is

Answer 1

Given, heat given in the process $A\to B,$

$Q_{A\to B}=56J$

From graph, $A\to B$ is isobaric process,

$Q_{A\to B}=n{C}_p\Delta T=n(\frac{fR}{2}+R)\Delta T$

$=\left(\frac{f+2}{2}\right)(3p_0{V}_0-p_o{V}_0)$

$=(f+2)p_0{V}_0\dots \left(i\right)$

Given, heat given in the process $A\to C,$

$Q_{A\to C}=360J$

For the process $A\to C$, using first law of thermodynamics,

$Q_{A\to C}=\Delta U+W$

$Q_{A\to C}=n{C}_V\Delta T+$ area under curve

$=n\frac{fR}{2}\Delta T+\frac{1}{2}\left(5P_0\right)\left(3V_0\right)$

$=\left(\frac{f+1}{2}\right)\left(15p_0{V}_0\right)\dots \left(ii\right)$

Dividing $\left(i\right)$ and $\left(ii\right)$, we get

$\frac{Q_{A\to B}}{Q_{A\to C}}=\frac{(f+2)2}{(f+1)15}=\frac{56}{360}$

$\Rightarrow f=5$

$\Rightarrow \gamma =\frac{f+2}{f}=1.40$

Final answer: $\left(1.40\right)$