Question

An ideal gas (γ = 1.67) is taken through the process abc shown in the figure. The temperature at point a is 300 K. Calculate (a) the temperatures at b and c (b) the work done in the process (c) the amount of heat supplied in the path ab and in the path bc and (d) the change in the internal energy of the gas in the process.

Answer 1

(a) For line ab, volume is constant.
So, from the ideal gas equation,
$$\begin{gathered} \frac{P_{\mathit{1}}}{T_{\mathit{1}}}\mathit{=}\frac{P_{\mathit{2}}}{T_{\mathit{2}}} \\ \Rightarrow \frac{100}{300}=\frac{200}{{\mathrm{T}}_2} \end{gathered}$$



$T_{\mathit{2}}=\left(\frac{200\times 300}{100}\right)=600\mathrm{K}$

For line bc, pressure is constant.
$$\begin{gathered} \mathrm{So},\frac{{\mathit{V}}_{\mathit{1}}}{{\mathit{T}}_{\mathit{1}}}\mathit{=}\frac{{\mathit{V}}_{\mathit{2}}}{{\mathit{T}}_{\mathit{2}}} \\ \Rightarrow \frac{100}{600}=\frac{150}{{\mathrm{T}}_2} \\ \Rightarrow {\mathrm{T}}_2\left(\frac{600\times 150}{100}\right)=900\mathrm{K} \end{gathered}$$

(b)
As process ab is isochoric, $W_{ab}=0$.
During process bc,
P = 200 kPa

The volume is changing from 100 to 150 cm³ .
Therefore, work done = 50 × 10⁻⁶ × 200 × 10³ J
= 10 J

(c) For ab (isochoric process), work done = 0.
From the first law,
dQ = dU = nC_vdT
$\Rightarrow$Heat supplied = nC_vdT
Now,
$$\begin{gathered} Q_{ab}\mathit{=}\left(\frac{PV}{RT}\right)\mathit{\times }\left(\frac{R}{\gamma \mathit{-}\mathit{1}}\right)\mathit{\times }dT \\ =\frac{200\times {10}^3\times 100\times {10}^{-6}\times 300}{600\times 0.67} \\ =14.925(\therefore \mathrm{\gamma }=1.67) \\ \text{F}\mathrm{or}\mathrm{bc}(\mathrm{isobaric}\mathrm{process}): \\ \mathrm{Heat}\mathrm{supplied}\mathrm{in}\mathrm{bc}=n{C}_{p}dT\mathit{}\mathit{}\mathit{}\mathit{}\left(C_p\mathit{=}\frac{\gamma R}{\gamma \mathit{-}\mathit{1}}\right) \\ \mathit{=}\frac{PV}{RT}\mathit{\times }\frac{\gamma R}{\gamma \mathit{-}\mathit{1}}\mathit{\times }dT \\ =\frac{200\times {10}^3\times 150\times {10}^{-6}}{600\times 0.67}\times 300 \\ =10\times \frac{1.67}{0.67}=\frac{16.7}{0.67}=24.925 \end{gathered}$$

(d) dQ = dU + dW
Now,
dU = dQ − dW
= Heat supplied − Work done
= (24.925 + 14.925) − 10
= 39.850 − 10 = 29.850 J.