91Ó°ÊÓ

Given,

Number of moles$n=0.450mol$

Temperature at point a :

From Ideal gas law we get:

$$\begin{gathered} PV=nRT \\ T=\frac{PV}{nR} \end{gathered}$$

From the figure we have that pressure at point $a,P=2×{10}^5\mathrm{Pa}$, Volume . Now putting these values in the above equation, we get:

$$\begin{gathered} T_a=\frac{pV}{nR} \\ =\frac{2×{10}^5×0.010m^3}{0.450mol×8.314J/molK} \\ =534.4K \end{gathered}$$

Similarly for point b , pressure$P=5×{10}^5$ and Volume$V=0.070m^3$ , therefore the temperature

$$\begin{gathered} T_b=\frac{5×{10}^{5}Pa×0.007m^3}{0.450mol×8.314J/mol∙K} \\ =9350K \end{gathered}$$

Again, for point c we have$P=8×{10}^5\mathrm{Pa}$ and Volume$V=0.070m^3$

Therefore, the temperature at$T_c$ is:

$$\begin{gathered} T_c=\frac{PV}{nR} \\ =\frac{8×{10}^{5}Pa×0.07m^3}{0.450mol×8.134J/molK} \\ =15030J \end{gathered}$$

Now, the process has two paths and to get the work done we can simply combine both the work done.

For the path ab pressure changes from$P_1=2×{10}^5\mathrm{Pa}$ to$P_2=5×{10}^5\mathrm{Pa}$ and the volume changes from$V_1=0.01m^3$ to$V_2=0.07m^3$ . Hence the work done is given by:

$\begin{array}{l}{W}_{ab}=\frac{1}{2}\left(p_1+p_2\right)\left(V_2-V_1\right)\\ =\frac{1}{2}\left(5×{10}^{5}Pa+2×{10}^{5}Pa\right)\left(0.07m^3+0.01m^3\right)\\ =21000J\end{array}$

For the process bc there is no change in volume hence the work done is zero, i.e. : $W_{bc}=0$.

Now the total work done for the entire process is:

$\begin{array}{l}{W}_t=W_{ab}+W_{bc}\\ =21000J+0\\ =21000J\end{array}$

Therefore, the total work done is$21000J$

To increase the internal energy by$∆U=15000J$ we can simply add heat. Now applying the first law of thermodynamics we get:

role="math" localid="1664278233217" $\begin{array}{l}Q=∆U+W\\ =15000+21000\\ =36000J\end{array}$

Therefore, the required heat needs to be added is 36000J to increase the internal energy by 15000J . The work done is positive.