Question

In: Physics

A 0.535-mol sample of an ideal diatomic gas at 408 kPa and 279 K expands quasi-statically...

A 0.535-mol sample of an ideal diatomic gas at 408 kPa and 279 K expands quasi-statically until the pressure decreases to 159 kPa. Find the final temperature and volume of the gas, the work done by the gas, and the heat absorbed by the gas if the expansion is the following.

(a) isothermal

final temperature	K
volume of the gas	L
work done by the gas	J
heat absorbed	J

(b) adiabatic

final temperature	K
volume of the gas	L
work done by the gas	J
heat absorbed	J

Expert Solution

initial pressure P1 = 408 kPa

initial temperature T1 = 279 K

nmber of moles n = 0.535 mol

initial volume V1 = n*R*T1/P1 = (0.535*8.314*279)/(408*10^3) = 0.00304 m^3

(a)

isothermal process temperatre is constant

P2 = 159 kPa

T2 = T1 = 29 K

V2 = P1*V1/P2 = (408*10^3*0.000304)/(159*10^3) = 0.00078 m^3

Work = 2.303*n*R*T1*log(P1/p2)

work = 2.303*0.535*8.314*279*log(408/159) = 1169.67 J

from Ist law of thermodynamics

heat absorbed Q = dU + W = 0 + 1169.67 J

==============

(b)

adiabatic ( no heat transfers )

ratio of specific heat = gamma = 1.4

P^(1-gamma)*T^gamma = constant

P1^(1-gamma)*T1^gamma = P2^(1-gamma)*T2^gamma

(408*10^3)^(1-1.4)*279^1.4 = (159*10^3)^(1-1.4)*T2^1.4

T2 = 213.14K

-------------
PV^gamma = constant

P1*V1^gamma = P2*V2^gamma

408*10^3*0.000304^1.4 = 159*10^3*v2^1.4

final volume V2 = 0.000595 m^3

===================

W = (P2*V2 - P1*V1)/(1-gamma)

W = ( (159*10^3*0.000595)-(408*10^3*0.000304) )/(1-1.4)

W = 73.6 J

Q = 0

genius_generous answered 2 years ago

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