Question

Starting at temperature 300 K, one mole of an ideal diatomic gas (ϒ = 1.4) is first compressed adiabatically from volume to V₁ to V₂ = V₁/16. It is then allowed to expand isobarically to volume 2V₂ . If all the processes are the quasi-static then the final temperature of the gas (⁰K) is (to the nearest integer).

Answer 1

Solution :-

Step : 1 Calculation of T₂ by using formula ${\mathrm{TV}}^{\mathrm{Y}}=\mathrm{Constant}$

$$\begin{gathered} \mathrm{Given},{\mathrm{T}}_1=300\mathrm{K} \\ {\mathrm{V}}_2=\frac{{\mathrm{V}}_1}{16} \\ {\mathrm{TV}}^{\mathrm{Y}}=\mathrm{Constant} \\ {\mathrm{T}}_1{\mathrm{V}}_1^{\mathrm{\gamma }}={\mathrm{T}}_2{\mathrm{V}}_2^{\mathrm{\gamma }} \\ 300{\left({\mathrm{V}}_1\right)}^{(1.4-1)}={\mathrm{T}}_2{\left(\frac{{\mathrm{V}}_1}{16}\right)}^{(1.4-1)} \\ 300{\left({\mathrm{V}}_1\right)}^{\left(\frac{2}{5}\right)}={\mathrm{T}}_2{\left(\frac{{\mathrm{V}}_1}{16}\right)}^{\left(\frac{2}{5}\right)} \\ {\mathrm{T}}_2=300\mathrm{x}{2}^{\left(\frac{8}{5}\right)} \end{gathered}$$

Step : 2 Calculation of Final temperature (Tc) from BC Process,

$$\begin{gathered} Given,T_2=300*2^{\left(\frac{8}{5}\right)} \\ \mathrm{Vc}=2{\mathrm{V}}_2 \\ \frac{{\mathrm{V}}_2}{{\mathrm{T}}_2}=\frac{Vc}{Tc} \\ Tc=\frac{Vc*T_2}{V_2} \\ Tc=\frac{2V_2*300*2^{\left(\frac{8}{5}\right)}}{V_2} \\ Tc=1818K \end{gathered}$$

Final answer : The final temperature of the gas (Tc) is 1818K.