A frictionless piston is free to move in a closed cylinder. Initially there is 0.035 m³ of oxygen at 4.5 bar, 60°C on one side of the piston and 0.07 m³ of methane at 4.5 bar and – 12°C on the other side. The cylinder walls and piston may be regarded as perfect thermal insulators but the oxygen may

Question

A frictionless piston is free to move in a closed cylinder. Initially there is 0.035 m³ of oxygen at 4.5 bar, 60°C on one side of the piston and 0.07 m³ of methane at 4.5 bar and – 12°C on the other side. The cylinder walls and piston may be regarded as perfect thermal insulators but the oxygen may be heated electrically. Heating takes place so that the volume of oxygen doubles. Find :

(i) Final state condition ;

(ii) Work done by the piston ;

(iii) Heat transferred to oxygen.

Treat both gases as perfect and take :

For oxygen $c_p=0.88$ kJ/kg K, R = 0.24 kJ/kg K

For methane $c_p=1.92$ kJ/kg K, R = 0.496 kJ/kg K.

(i) Final state condition ;

(ii) Work done by the piston ;

(iii) Heat transferred to oxygen.

Treat both gases as perfect and take :

For oxygen [latex]c_p=0.88[/latex] kJ/kg K, R = 0.24 kJ/kg K

For methane [latex]c_p=1.92[/latex] kJ/kg K, R = 0.496 kJ/kg K.

Accepted Answer

For oxygen :

Initial volume, [latex]V_1=0.035[/latex] m³

Initial pressure, [latex]p_1=4.5[/latex] bar

Initial temperature, [latex]T_1=60+273=333[/latex] K

For methane :

Initial volume, [latex]V_1=0.07[/latex] m³

Final volume, [latex]V_2=0.035[/latex] m³

Initial pressure, [latex]p_1=4.5[/latex] bar

Initial temperature of methane,

[latex]T_1=-12+273=261[/latex] K.

For Methane :

[latex]c_p=R imes \frac{\gamma}{\gamma-1} \quad ext { or } \quad 1.92=0.496\left(\frac{\gamma}{\gamma-1} ight)[/latex]

or [latex]\frac{1.92}{0.496}=\frac{\gamma}{\gamma-1} \quad ext { or } \quad 1.92(\gamma-1)=0.496 \gamma[/latex]

∴ [latex]\gamma=\frac{1.92}{(1.92-0.496)}=1.348 ext { say } 1.35[/latex]

For Oxygen :

[latex]c_v=c_p-R=0.88-0.24=0.64[/latex] kJ/kg K.

(i) According to problem ; for methane

[latex]p V^\gamma=[/latex] constant holds good

∴ [latex]p_1 V_1^\gamma=p_2 V_2^\gamma[/latex]

[latex]p_2=p_1 \cdot\left(\frac{V_1}{V_2} ight)^\gamma=4.5(2)^{1.35}=11.47[/latex] bar.

Also, [latex]\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}[/latex]

or [latex]T_2=\frac{p_2 V_2 T_1}{p_1 V_1}=\frac{11.47 imes 0.035 imes 261}{4.5 imes 0.07}=332.6[/latex] K.

∴ [latex] ext { Work done }=\frac{p_1 V_1-p_2 V_2}{\gamma-1}=\frac{4.5 imes 10^5 imes 0.07-11.47 imes 10^5 imes 0.035}{(1.35-1)}[/latex] J

[latex] ext { Work done }=\frac{p_1 V_1-p_2 V_2}{\gamma-1}=\frac{4.5 imes 10^5 imes 0.07-11.47 imes 10^5 imes 0.035}{(1.35-1)}[/latex] kJ

= – 24.7 kJ (done on the methane)

(ii) The piston will be in virtual equilibrium and hence zero work is effected by the piston.

(iii) Work done by oxygen = work done on methane and expansion of oxygen is effected in the system

∴ [latex]W_{ ext {oxygen }}=+24.7[/latex] kJ

and [latex]Q=\left(U_2-U_1 ight)+W[/latex]

Amount of oxygen present [latex]=\frac{p_1 V_1}{R T_1}=\frac{4.5 imes 10^5 imes 0.035}{0.24 imes 1000 imes 333}=0.197[/latex] kg

and [latex]T_2=\frac{p_2 V_2}{p_1 V_1} imes T_1=\frac{11.47 imes 0.07 imes 333}{4.5 imes 0.035}= 1 6 9 7 . 5[/latex] K.

(As the piston is free, the final pressure of oxygen and methane will be same).

∴ [latex]Q=\left(U_2-U_1 ight)+W[/latex]

= [latex]m c_v\left(T_2-T_1 ight)+W[/latex]

= 0.197 × 0.64 (1697.5 – 333) + 24.7 = 196.7 kJ.

Question 4.66: A frictionless piston is free to move in a closed cylinder. ...

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