<span>Note that internal energy is a state function. That means internal energy of the gas can be expressed as function of two state variables, e.g. U = f(T;V). For an ideal gas internal function can be expressed of temperature alone. But is not necessary to make ideal gas assumption to solve this problem.
Because internal energy is a state function, a process changing from state 1 to state 2 has always the same change change in internal energy irrespective of the process design.
The one-step compression and the two two-step compression start at the sam state and end up in same state. the gas undergoes the same change in internal energy:
∆U₁ = ∆U₂
The change in internal energy of the gas equals the heat added to the gas plus work done on it:
Hence,
Q₁ + W₁ = Q₂ + W₂
So the difference in heat transfer between the two process is:
∆Q = Q₂ - Q₁ = W₁ - W₂
The work done on the gas is given by piston is given by the integral
W = - ∫ P_ex dV from V_initial to V_final
For constant external pressure like in this problem this simplifies to
W = - P_ex ∙ ∫ dV from V_initial to V_final
= P_ex ∙ (V_initial - V_final)
I hope my guide has come to your help. Have a nice day ahead and may God bless you always!</span>
Answer:
I think the SI unit of A is m^2/s
Answer:
1) 
2) 
3) 
4) The Molecules do not burn because of the presences of hydrogen bond in place
Explanation:
From the question we are told that
latent heat of vaporization for water at room temperature is 2430 J/g.
1)Generally in determining the molar mass of water evaporated we have that
-One mole (6.02 x 10. 23 molecules)
-Molar mass of water is 18.02 g/mol
Mathematically the mass of water is give as
Therefore
b)Generally the evaporation speed V is given as
Mathematically derived from the equation
To Give
c)Generally the equation for velocity 
Therefore
Effective temperature T is given by
where
4) The Molecules do not burn because of the presences of hydrogen bond in place
