Respuesta :
Answer:
a) W = 900 J. b) Q = 3142.8 J . c) ΔU = 2242.8 J. d) W = 0. e) Q = 2244.78 J. g) Δ U = 0.
Explanation:
(a) Work done by the gas during the initial expansion:
The work done W for a thermodynamic constant pressure process is given as;
W = p Δ V
where
p is the pressure and Δ V is the change in volume.
Here, Given;
P 1 = i n i t i a l p r e s s u r e = 2.5 × 10^ 5 P a
T 1 = i n i t i a l t e m p e r a t u r e = 360 K
n = n u m b er o f m o l e s = 0.300 m o l
The ideal gas equation is given by
P V = nRT
where ,
p = absolute pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = universal gas constant = 8.314 K J / m o l K
T = absolute temperature of the gas
Now we will Calculate the initial volume of the gas using the above equation as follows;
PV = n R T
2.5 × 10 ^5 × V 1 = 0.3 × 8.314 × 360
V1 = 897.91 / 250000
V 1 = 0.0036 m ^3 = 3.6×10^-3 m^3
We are also given that
V 2 = 2× V 1
V2 = 2 × 0.0036
V2 = 0.0072 m^3
Thus, work done is calculated as;
W = p Δ V = p×(V2 - V1)
W = ( 2.5 × 10 ^5 ) ×( 0.0072 − 0.0036 )
W = 900 J.
(b) Heat added to the gas during the initial expansion:
For a diatomic gas,
C p = 7 /2 ×R
Cp = 7 /2 × 8.314
Cp = 29.1 J / mo l K
For a constant pressure process,
T 2 /T 1 = V 2 /V 1
T 2 = V 2 /V 1 × T 1
T 2 = 2 × T 1 = 2×360
T 2 = 720 K
Heat added (Q) can be calculated as;
Q = n C p Δ T = nC×(T2 - T1)
Q = 0.3 × 29.1 × ( 720 − 360 )
Q = 3142.8 J .
(c) Internal-energy change of the gas during the initial expansion:
From first law of thermodynamics ;
Q = Δ U + W
where ,
Q is the heat added or extracted,
Δ U is the change in internal energy,
W is the work done on or by the system.
Put the previously calculated values of Q and W in the above formula to calculate Δ U as;
Δ U = Q − W
ΔU = 3142.8 − 900
ΔU = 2242.8 J.
(d) The work done during the final cooling:
The final cooling is a constant volume or isochoric process. There is no change in volume and thus the work done is zero.
(e) Heat added during the final cooling:
The final process is a isochoric process and for this, the first law equation becomes ,
Q = Δ U
The molar specific heat at constant volume is given as;
C v = 5 /2 ×R
Cv = 5 /2 × 8.314
Cv = 20.785 J / m o l K
The change in internal energy and thus the heat added can be calculated as;
Q = Δ U = n C v Δ T
Q = 0.3 × 20.785 × ( 720 - 360 )
Q = 2244.78 J.
(f) Internal-energy change during the final cooling:
Internal-energy change during the final cooling is equal to the heat added during the final cooling Q = Δ U .
(g) The internal-energy change during the isothermal compression:
For isothermal compression,
Δ U = n C v Δ T
As their is no change in temperature for isothermal compression,
Δ T = 0 , then,
Δ U = 0.
