A gas is confined to a vertical cylinder by a piston of mass 2 kg and radius 1 cm. When 5J of heat are added, the piston rises b
1 answer:
The work done by gas is 0.753 J and change in internal energy is 4.247J
So we are given that mass is 2kg , radius 1 cm and the amount of heat is 5 cm
The piston raised by 2.4cm
As we know that Work done is PΔV
Where ΔV is change in volume
Therefore ΔV = πr^2 h = π x (.01)^2 x .024 =7.53×10^(-6)m^3
Here pressure is 10^5 pa
So W =
Therefore W = 0.753 J
Now coming to change in internal energy
Change in Internal Energy = Heat Added - Energy lost in work
∴ 5J - 0.753 J = 4.247J
Hence the change in internal energy is 4.247 J
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Answer : The change in enthalpy of the reaction is, -310 kJ
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The given main reaction is,
The intermediate balanced chemical reaction will be,
(1)
(2)
(3)
Now we will reverse the reaction 1 and multiply reaction 1 by 2, reaction 2 by 2 and reaction 3 by 3 then adding all the equations, we get :
(1)
(2)
(3)
The expression for enthalpy of formation of will be,
Therefore, the change in enthalpy of the reaction is, -310 kJ