In this problem you are to consider an adiabaticexpansion of an ideal diatomic gas, which means that the gas expands with no add
ition or subtraction of heat.
Assume that the gas is initially at pressure p0, volume V0, and temperature T0. In addition, assume that the temperature of the gas is such that you can neglect vibrational degrees of freedom. Thus, the ratio of heat capacities is γ=Cp/CV=7/5.
Note that, unless explicitly stated, the variable γ should not appear in your answers--if needed use the fact that γ=7/5 for an ideal diatomic gas.
Part A
Find an analytic expression for p(V), the pressure as a function of volume, during the adiabatic expansion.
Express the pressure in terms of V and any or all of the given initial values p0, T0, and V0.
Correct
p(V) =
p0(V0V)75
Part B
At the end of the adiabatic expansion, the gas fills a new volume V1, where V1>V0. Find W, the work done by the gas on the container during the expansion.
Express the work in terms of p0, V0, and V1. Your answer should not depend on temperature.
Part C
Find ΔU, the change of internal energy of the gas during the adiabatic expansion from volume V0 to volume V1.
Express the change of internal energy in terms of p0, V0, and/or V1.
Need Part B and C
Assume that the gas is initially at pressure p0, volume V0, and temperature T0. In addition, assume that the temperature of the gas is such that you can neglect vibrational degrees of freedom. Thus, the ratio of heat capacities is γ=Cp/CV=7/5.
Note that, unless explicitly stated, the variable γ should not appear in your answers--if needed use the fact that γ=7/5 for an ideal diatomic gas.
Part A
Find an analytic expression for p(V), the pressure as a function of volume, during the adiabatic expansion.
Express the pressure in terms of V and any or all of the given initial values p0, T0, and V0.
Correct
p(V) =
p0(V0V)75
Part B
At the end of the adiabatic expansion, the gas fills a new volume V1, where V1>V0. Find W, the work done by the gas on the container during the expansion.
Express the work in terms of p0, V0, and V1. Your answer should not depend on temperature.
Part C
Find ΔU, the change of internal energy of the gas during the adiabatic expansion from volume V0 to volume V1.
Express the change of internal energy in terms of p0, V0, and/or V1.
Need Part B and C
1 answer:
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Answer:
155.38424 K
2.2721 kg/m³
Explanation:
= Reservoir pressure = 10 atm
= Reservoir temperature = 300 K
= Exit pressure = 1 atm
= Exit temperature
= Specific gas constant = 287 J/kgK
= Specific heat ratio = 1.4 for air
Assuming isentropic flow
Flow temperature at exit is 155.38424 K
Density at exit can be derived using the ideal gas equation
Flow density at exit measures 2.2721 kg/m³
Answer:
a, 71.8° C, 51° C
b, 191.8° C
Explanation:
Given the data:
D(i) = 200 mm
D(o) = 400 mm
q' = 24000 W/m³
k(r) = 0.5 W/m.K
k(s) = 4 W/m.K
k(h) = 25 W/m².K
The heat generation formula can be articulated as follows:
q = πr²Lq'
q = π. 0.1². L. 24000
q = 754L W/m
Thermal conduction resistance, R(cond) = 0.0276/L
Thermal conduction resistance, R(conv) = 0.0318/L
Applying the energy balance equation,
Energy In = Energy Out
This equates to q, which is 754L
From the initial analysis, the temperature at the interface between the rod and sleeve is found to be 71.8° C
Additionally, the outer surface temperature records as 51° C
Furthermore, based on the second analysis, the calculated temperature at the center of the rod is determined to be 191.8° C
Para calcular el peso utilizamos la fórmula:
Cuando tenemos dos cuerpos, empleamos la fórmula de la gravedad general:
Donde:
G = constante de gravedad =
Mp = masa de la persona = 882 / 9.81= 89.9kg
ME = masa de la Tierra =
r = distancia entre la Tierra y la persona
A partir de estas dos fórmulas, deducimos que los lados izquierdos son equivalentes, por lo tanto, los lados derechos también deben serlo.
Resolviendo esta ecuación para r:
r=6371116m = 6371.116km
Esta es la distancia desde el centro de la Tierra. El radio de la Tierra es 6370km y la altura sobre la superficie es 6371.116 - 6370 = 1.116km o 1116m.
Response:
Clarification:
We need an expression that shows how much water has been drained from the tub. This is represented by v, which indicates how many gallons have flowed out since the plug was taken out. Each gallon removed equates to 8.345 pounds of water, so the weight of the drained water Q in pounds as a function of v can be expressed as:
Where v signifies the number of gallons emptied from the tub.
Have a great day! Let me know if there's anything else I can assist with.