A 3.75 mol sample of a diatomic ideal gas ( ? = 1.4) expands slowly and adiabati

Question

A 3.75 mol sample of a diatomic ideal gas (? = 1.4) expands slowly and adiabatically from a pressure of 4.25 atm and a volume of 13.0 L to a final volume of 33.0 L.

A 3.75 mol sample of a diatomic ideal gas (? = 1.4) expands slowly and adiabatically from a pressure of 4.25 atm and a volume of 13.0 L to a final volume of 33.0 L. What is the final pressure of the gas? What is the initial temperature of the gas? What is the final temperature of the gas? Find the value for Q. Find the value for ?U. Find the value for W (work done by the gas). Use ?U = Q - W and the answers from the last two parts.

Explanation / Answer

(P1V1)^ ? =(P2V2) ^?
where p- pressure, atm
V-volume, Litre
?- 1.4 adiabatic exponent

(4.25*13)^1.4 = (P2*33)^1.4

------>P2= 1.674 atm

We know,
out of this you can get the initial temperature & final temperatures
pV= mRT
m- number of mols
T-temperature

Initial temperature, T1 = (4.25*13)/3.75*8.314

                                     = 1.77 C

                         Final, T2 = (33*1.674)/8.314*3.75

------> Q is always 0 in an adiabatic process
-------> L=?U= ?*2.5*R*(T2-T1)
-------->Eint= ?U
R- the constant of ideal gases

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