The internal energy of a system changes because the system gains 160 j of heat and performs 309 j of work. in returning to its initial state, the system loses 106 j of heat. during this return process, what work is involved, and is the work done by the system or on the system?

w=255

Explanation:

The change in internal energy is given by the first law:

ΔE = Q - w

where ΔE is the change in internal energy of the system

q is the heat added to the system

w is the work done *by* the system on the surroundings

So, for the first phase of this process:

ΔE = Q - w

Q=160J

w=309J

ΔE = 160J - 309J = -149J

To bring the system back to its initial state after this, the internal energy must change by +149J (the system myst gain back the 149 J of energy it lost).  We are told that the system loses 106 J of heat in returning to its initial state, so the work involved is given by:

ΔE = Q - w

+149J = -106J - w

255J = -w

w = -255J

its c its is a mixture of gases

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂