A tank of capacity 0.45 m³ is insulated and is divided into two sections through a partition. One section initially contains H2 at 3 bar and 130°C and has a volume of 0.3 m³ and the other section initially holds N2 at 6 bar and 30°C. The gases are then allowed to mix after removing the adiabatic

Question

A tank of capacity 0.45 m³ is insulated and is divided into two sections through a partition. One section initially contains [latex]H_2[/latex] at 3 bar and 130°C and has a volume of 0.3 m³ and the other section initially holds [latex]N_2[/latex] at 6 bar and 30°C. The gases are then allowed to mix after removing the adiabatic partition. Determine :

(i) The temperature of the equilibrium mixture ;

(ii) The pressure of the mixture ;

(iii) The change in entropy for each component and total value.

Assume : [latex]c_{v_{\left(N_2\right)}}=0.744[/latex] kJ/kg K, [latex]c_{v_{\left(H_2\right)}}=10.352[/latex] kJ/kg K

[latex]c_{p_{\left(N_2\right)}}=1.041[/latex] kJ/kg K, [latex]c_{p_{\left(H_2\right)}}=14.476[/latex] kJ/kg K.

Accepted Answer

Total capacity of the tank, V = 0.45 m³

[latex]V_{ H _2}=0.3[/latex] m³ ; [latex]T_{ H _2}=130+273=403[/latex] K

[latex]p_{ H _2}=3[/latex] bar ; [latex]V_{ N _2}=0.15[/latex] m³ (i.e., 0.45 – 0.3 = 0.15 m³)

[latex]p_{ N _2}=6[/latex] bar ; [latex]T_{ N _2}=30+273=303[/latex] K.

(i) Temperature of equilibrium mixture, [latex]T _2[/latex] :

Now [latex]p_{ H _2} V_{ H _2}=m_{ H _2} R_{ H _2} T_{ H _2}[/latex]

∴ [latex]m_{ H _2}=\frac{3 imes 10^5 imes 0.3}{\left(\frac{8.314}{2} ight) imes 403 imes 10^3}=0.0537[/latex] kg [latex]\left[\because R=\frac{R_0}{M} ight][/latex]

and [latex]p_{ N _2} V_{ N _2}=m_{ N _2} R_{ N _2} T_{ N _2}[/latex]

∴ [latex]m_{ N _2}=\frac{6 imes 10^5 imes 0.15}{\left(\frac{8.314}{28} ight) imes 303 imes 10^3}=1.0[/latex] kg

According to the first law for a closed system

[latex]\Delta U=0[/latex]

i.e., [latex]U_{ ext {initial }}=U_{ ext {final }}[/latex]

or [latex]m_{ H _2} c_{v_{\left( H _2 ight)}} T_{\left( H _2 ight)_1}+m_{ N _2} c_{v_{\left( N _2 ight)}} T_{\left( N _2 ight)_1}=m_{ H _2} c_{v_{\left( H _2 ight)}} T_{\left( H _2 ight)_2}+m_{ N _2} c_{v_{\left( N _2 ight)}} T_{\left( N _2 ight)_2}[/latex]

or [latex]m_{ H _2} c_{v_{\left( H _2 ight)}}\left[T_{\left( H _2 ight)_2}-T_{\left( H _2 ight)_1} ight]+m_{ N _2} c_{v_{\left( N _2 ight)}}\left[T_{\left( N _2 ight)_2}-T_{\left( N _2 ight)_1} ight]=0[/latex]

or [latex]0.0537 imes 10.352\left[T_{\left( H _2 ight)_2}-403 ight]+1.0 imes 0.744\left[T_{\left( N _2 ight)_2}-303 ight]=0[/latex]

[latex]T_{\left( H _2 ight)_2}=T_{\left( N _2 ight)_2}=T_2[/latex]

or [latex]0.556\left(T_2-403 ight)+0.744\left(T_2-303 ight)=0[/latex]

or [latex]0.556 T_2-224+0.744 T_2-225.4=0[/latex]

∴ [latex]T_2=345.7[/latex] K

i.e., Temperature of the mixture = 345.7 K.

(ii) Pressure of the mixture, [latex]p _2[/latex] :

Now [latex]p_{\left( H _2 ight)_2} V=m_{ H _2} R_{ H _2} T_2[/latex]

∴ [latex]p_{\left( H _2 ight)_2}=\frac{0.0537 imes\left(\frac{8.314}{2} ight) imes 345.7 imes 10^3}{0.45 imes 10^5}=1.71[/latex] bar

Similarly [latex]p_{\left( N _2 ight)_2}=\frac{1.0 imes\left(\frac{8.314}{28} ight) imes 345.7 imes 10^3}{0.45 imes 10^5}=2.28[/latex] bar

∴ [latex]p _2=p_{\left( H _2 ight)_2}+p_{\left( N _2 ight)_2}=1.71+2.28= 3 . 9 9[/latex] bar.

(iii) Change in entropy :

Now [latex](\Delta S)_{ H _2}=m\left(c_p \log _e \frac{T_2}{T_1}-R \log _e \frac{p_2}{p_1} ight)_{ H _2}[/latex]

= [latex]0.0537\left[14.476 \log _e \frac{345.7}{403}-\frac{8.314}{2} \log _e \frac{1.71}{3} ight][/latex]

= 0.00626 kJ/K.

[latex](\Delta S)_{ H _2}=m\left(c_p \log _e \frac{T_2}{T_1}-R \log _e \frac{p_2}{p_1} ight)[/latex]

= [latex]1.0\left[1.041 \log _e \frac{345.7}{303}-\frac{8.314}{28} \log _e \frac{2.28}{6} ight][/latex]

= 0.424 kJ/K.

∴ [latex]\Delta S=(\Delta S)_{ H _2}+(\Delta S)_{ N _2}=0.00626+0.424=0.43026[/latex] kJ/K

i.e., Total change in entropy = 0.43026 kJ/K.

Question 9.21: A tank of capacity 0.45 m³ is insulated and is divided into ...

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