A rigid and insulated cylinder is divided into two compartments by a movable pis

Question

A rigid and insulated cylinder is divided into two compartments by a movable piston with low thermal resistance. Compartment A initially contains 80 kg of oxygen gas (02) at a temperature and pressure of 300 K and 100 kPa. Compartment B initially contains 20 kg of helium gas at 1200 K and 100 kPa. Internal heat transfer slowly occurs across the piston from the hot helium gas to the colder oxygen gas. The piston moves due to the changing temperatures, pressures and volumes. Finally thermal equilibrium will be reached to stop the process. Treat the helium and oxygen as a single system so that overall system has no external heat transfer (insulated) and no external work (rigid cylinder with constant total volume). Mass & energy must balance (MEB) so balance the mass and energy equations to solve. Use the perfect gas constants from the class equation sheet. The masses of the gases is given. With total volume known, and final temperature known, and the final state in thermal and mechanical equilibrium there is a single final temperature and throughout the process the pressure is equal for both gases (the piston is free to move). The final pressure of both gases (to the nearest integer value in kPa) is approximately.

Explanation / Answer

Step 1 find the volume occupied by helium and oxygen using ideal gas equation as they follow perfect gas equation

let Va = volume occupied by helium at the beginning = (mh * Rh * Th) /Pi

mh =mass of helium present = 20kg

Rh =gas constant of helium =Universal gas constant/ Molecular weight =8.314/4 kJ/kg K

Th = initial helium temperature = 1200K

Pi= initial pressure = 100kPa

Va= (mh * Rh * Th) /Pi = 498.8m3

let Vb = volume occupied by helium at the beginning = (mo * Ro * To) /Pi

mo =mass of oxygen present = 80kg

Ro =gas constant of oxygen =Universal gas constant/ Molecular weight =8.314/16 kJ/kg K

To = initial oxygen temperature = 300K

Pi= initial pressure = 100kPa

Vb = (mo * Ro * To) /Pi = 124.7 m3

Total volume V = Va +Vb = 623.5 m3

Let final temperature of two gases = Tf K

Volume occupied by helium after cooling= Vc

Volume occupied by oxygen after heating= Vd

We know pressures are equal at every instant so at the end of cooling of helium equate pressures Pf

Pf = (mh * Rh * Tf) /Vc = (mo * Ro * To) / Vd

After substituting values we get   Vc = Vd

So Vc =Vd = V/2= 623.5/2 =311.75m3

Now find final temperature using one of the equations of energy conservation

(mh * Rh * (Th - Tf)) =(mo * Ro * (Tf -To))

Tf = 750 K

Now find pressure using one of the equations = Pf = (mh * Rh * Tf) /Vc

                                                                                                                                                                                                                                                     = (20 *(8.314/4) * 750)/ 311.75

                                                            Final pressure = 100.00kPa

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