Two insulated tanks are connected through a pipe with closed valve in between. Initially one tank having volume of 1.8 m^3 has argon gas at 12 bar, 40°C and other tank having volume of 3.6 m^3 is completely empty. Subsequently valve is opened and the argon pressure gets equalized in two tanks. Dete

Question

Two insulated tanks are connected through a pipe with closed valve in between. Initially one tank having volume of 1.8 [latex] extbf{m}^{3}[/latex] has argon gas at 12 bar, 40°C and other tank having volume of 3.6 [latex] extbf{m}^{3}[/latex] is completely empty. Subsequently valve is opened and the argon pressure gets equalized in two tanks. Determine, (a) the final pressure & temperature (b) the change of enthalpy and (c) the work done considering argon as perfect gas and gas constant as 0.208 kJ/kg. K

Accepted Answer

[latex]\begin{aligned} ext{Total volume, } V = V_{1} + V_{2}&= 5.4 m^{3} \ ext{By perfect gas law, }p_{1}V_{1}&= mRT_{1}\12 × 10^{2} × 1.8 = m × 0.208 &× 313\⇒ \qquad\quad\qquad \qquad \qquad m&= 33.18 kg\end{aligned}\ ext{By gas law for initial and final state,} \\begin{aligned}\qquad\qquad\qquad \qquad p_{1}V_{1}&= p_{final} × V_{final}\12 × 10^{2} × 1.8&= p_{final} × 5.4\ ext{Final pressure} ⇒ p_{final}&= ext{400 kPa or 4 bar}\end{aligned}\ ext{Here since it is insulated system and it has no heat transfer so, there will be no change in internal energy,}\ ext{hence there will be no change in temperature. Also by Ist law of thermodynamics, since there is no}\ ext{heat transfer due to system being insulated and no work due to frictionless expansion;}\\qquad ext{Final temperature = 313 K.}\ \begin{aligned}dq&= du + dw ⇒ du = 0\i.e. \qquad \qquad \qquad T_{initial}&= T_{final}\ ext{Change in enthalpy}&= 0\ ext{Work done}&= 0\end{aligned}[/latex]

Question 4.19: Two insulated tanks are connected through a pipe with closed...

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