NOTE: The problems are all subparts of each other. I noted the numbers I was abl
ID: 2037479 • Letter: N
Question
NOTE: The problems are all subparts of each other. I noted the numbers I was able to find in parenthesis. I just need help with the bolded questions.
Suppose 2 moles of a monatomic ideal gas occupies 5 m^3 at a pressure of 1600 Pa. Find the temperature of the gas in Kelvin? (I found it 481 K) Find the total internal energy of the gas? (I found it 12000 J ). Suppose the gas sundergoes an isobaric expansion to a volume of 7 m^3. Find Q? Find W? (I got -3200 J) Find U? (I got 4800 J). Suppose the gas then undergoes adiabatic expansion to a pressure of 900 Pa and a volume of 11.5 m3. Find W? Find ?U? The gas then undergoes isothermal compression to a volume of 5 m3. Find Q? Find W? The gas then undergoes an isochoric process and returns to 1600 Pa of pressure. Find Q? Find ?U ?
Explanation / Answer
(A) P V = n R T
(1600) (5) = (2)(8.314)(T)
T = 481 K
(B) U = 3 n R T / 2 OR 3 P V / 2
= (1.5)(1600)(5)
= 12000 J
--------------------------
For isobaric,
W = P deltaV = 1600 x (7 - 5) = 3200 J ....Ans
Q = n Cp deltaT = (5/2)(n R deltaT ) = (2.5) P deltaV
Q = 8000 J .....And
delta(U) = Q - W = 8000 - 3200 = 4800 J
Uf - Ui = 4800
U = (4800) + 12000 = 16800 J .....Ams
----------------------------------
Adiabatic:
W = (900)(11.5)^(5/3) [ 11.5^(--2/3) - 7^(-2/3)] / (-2/3)
W = 6090 J .....Ans
deltaU = -W =- 6090 J .....Ans
-----------------------
Isothermal:
W = P V ln(VF/Vi) = (900 x 11.5) ln(5/11.5)
W = - 8620 J ....Ans
Q = W = - 8620 J .....Ans
----------------------
Isochoric:
Q = deltaU = n Cv deltaT = (3/2)(n R deltaT)
= (3/2) V deltaP
= (3/2) (5)(1600 - 2070)
= - 3525 J ....Ans
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.