91Ó°ÊÓ

Given,

It is given an ideal gas with moles$n=0.0175mol$ From the graph we can see that at point the pressure$p=0.2atm$ and the volume is$V=2L$

Now, applying the ideal gas formula and solve for T we get:

$T=\frac{pV}{nR}$, where$R$ is gas constant equals to$8.314J$ and the pressure is converted to Pa that is$1.013×{10}^{5}Pa$

$$\begin{gathered} =\frac{\left(0.2atm×1.013×{10}^{5}Pa\right)\left(2×{10}^{-3}{m}^3\right)}{0.0175mol×8.314J/mol∙K} \\ =278K \end{gathered}$$

Therefore, the lowest temperature isdata-custom-editor="chemistry" $278\mathrm{K}$

Since in the process from a to b the volume is constant therefore the work done is $W_{ab}=0$

Now, in the process from b to c the pressure decreases from 0.5atmto 0.3 atmand the volume increases from 2L to 6L .

Therefore, the work done from b to c is given by:

$$\begin{gathered} W_{ab}=\frac{1}{2}\left[\left(0.3+0.5\right)\left(1.013×{10}^{5}Pa\right)\right]×[\left(6-2\right)×{10}^{-3}{m}^3] \\ =162J \end{gathered}$$

Therefore, the work done is$162J$

Now it is given the value to heat added$Q=215J$ . So, to calculate the heat added we use the first law of thermodynamics

$$\begin{gathered} ∆U=Q-W \\ =215J-162J \\ =53J \end{gathered}$$

Therefore, the heat absorbed is$53J$