An ideal gas (c}r), flowing at 4 kmol/h, expands isothermally at 475 kfrom 100 to 50 kpa through a rigid device. if the power produced is 2000 w, what is the corresponding rate of heat flow? what is the rate of lost work?

The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

Explanation:

We are given:

Rate of flow of ideal gas , n = 4 kmol/hr =    (Conversion factors used:  1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW     (Conversion factor:  1 kW = 1000 W)

We know that:

  (For isothermal process)

So, by applying first law of thermodynamics:

     .......(1)

Now, calculating the work done for isothermal process, we use the equation:

where,

= change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

= initial pressure = 100 kPa

= final pressure = 50 kPa

Putting values in above equation, we get:

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

thats a little if u want the answer just look it up on google thats what i do haha : )