A closed, rigid, insulated vertical cylinder is divided into two compartments by

Question

A closed, rigid, insulated vertical cylinder is divided into two compartments by a metal pison of area A=0.12m^2; it contains 200 mol of CO2 in the top and 350 mol in the bottom compartment. The internal useful volume of the cylinder is V=1m^3 (net of piston volume). The piston can slide without friction and has mass m=150 kg. At the initial state, the temperature in both compartments is T=320 K. a.) What is the initial pressure in each of the two compartments? b.) An 5 kW electrical heater inside the cylinder is now turned on for 6 minutes. Obtain the common final temperature, and the pressure in each of the two compartments after a new equilibrium state has been reached. (Assume CO2 as ideal gas)

Explanation / Answer

number of moles in ist compartment = n1 = 200mol

number of moles in second compartment = n2=350mol

Temperature in both compartment = T = 320K

Volume of each compartment = 0.5m3

Using ideal gas equation

PV = nRT

P = nRT / V

For the compartment haing n=200mol , the intial pressure will be

P = 200(8.13) (320) / 0.5

P = 1040640 Pa

For the compartment having n = 350mol the inital pressure will be

P = (350)(8.13) (320) / 0.5

P = 1821120 pa

This will be the pressure in the bottom compartment

b) we know that

Q = mCv dT

Cv = specific heat capacity for carbon dioxide at constant volume and its value is

Cv =0.655 KJ / K kg

Q = P t =( 5 KW) (6)(60)sec

Q = 1800KJ

Now

Q = mCv ( T2-T1)

T2 = T1+ mQ / Cv

T2 = 320K + 1800KJ(8.8Kg) / o.655 KJ/K Kg ( weight in gram of CO2 in ist compartment =( 200g) ( 44g / mol) = 8800g T2 =24503 K

This would be the final temperature at the top compartment

T2= 320K + 1800KJ(15.4Kg) / o.655 KJ/K Kg

T2 = 42640K

This would be the final temperature at the bottom compartment

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