When the vessel and its contents are warmed to 100 °C, Q decomposes into its constituent elements. What is the total pressure, a
nd what are the partial pressures of xenon and oxygen in the container?
1 answer:
Answer:
The question is incomplete, but here is a similar question with complete details ; Gaseous compound Q contains only xenon and oxygen . When 0.100 g of Q is placed in a 50.0 mL steel vessel at 0 °C, the pressure is 0.229 atm. (a) What is the molar mass of Q (b) When the vessel and its contents are warmed to 100 °C, Q decomposes into its constituent elements . What is the total pressure , and what are the partial pressures of Xenon and Oxygen
Explanation:
The step by step calculations is as shown in the attached file.
It should be noted that what was applied is
- The Ideal gas equation PV = nRT
- Daltons law of partial pressure which states that in a mixture of gases, the total pressure exerted is equal to the sum of the individual partial pressures of the gases at constant temperature.
- It should be noted that the total pressure of the gases can be gotten by applying pressures law at constant volume
- P1/P2 = T1/T2
- It should also be noted that Partial pressure = Total pressure x Mole fraction
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Answer:
A) W' = 178.568 KW
B) ΔS = 2.6367 KW/k
C) η = 0.3
Explanation:
We are given;
Temperature at state 1;T1 = 360 °C
Temperature at state 2;T2 = 160 °C
Pressure at state 1;P1 = 10 bar
Pressure at State 2;P2 = 1 bar
Volumetric flow rate;V' = 0.8 m³/s
A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;
Specific volume;v1 = 0.287322 m³/kg
Mass flow rate of water vapour at turbine is defined by the formula;
m' = V'/v1
So; m' = 0.8/0.287322
m' = 2.784 kg/s
Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;
Specific enthalpy;h1 = 3179.46 KJ/kg
Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;
Specific enthalpy;h2 = 3115.32 KJ/kg
Now, since stray heat transfer is neglected at turbine, we have;
-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]
Potential and kinetic energy can be neglected and so we have;
-W' = m'(h2 - h1)
Plugging in relevant values, the work of the turbine is;
W' = -2.784(3115.32 - 3179.46)
W' = 178.568 KW
B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;
Specific entropy: s1 = 7.3357 KJ/Kg.k
Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;
Specific entropy; s2 = 8.2828 KJ/kg.k
The amount of entropy produced is defined by;
ΔS = m'(s2 - s1)
ΔS = 2.784(8.2828 - 7.3357)
ΔS = 2.6367 KW/k
C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;
h2s = 2966.14 KJ/Kg
Energy equation for turbine at ideal process is defined as;
Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]
Again, Potential and kinetic energy can be neglected and so we have;
-W' = m'(h2s - h1)
W' = -2.784(2966.14 - 3179.46)
W' = 593.88 KW
the isentropic turbine efficiency is defined as;
η = W_actual/W_ideal
η = 178.568/593.88 = 0.3
Answer:
Amount of air left in the cylinder=m=0.357 Kg
The amount of heat transfer=Q=0
Explanation:
Given
Initial pressure=P1=300 KPa
Initial volume=V1=0.2
Initial temperature=T=20 C
Final Volume==0.1
Using gas equation
m1==(300*0.2)/(.287*293)
m1=0.714 Kg
Similarly
m2=(P2*V2)/R*T2
m2=(300*0.1)/(0.287*293)
m2=0.357 Kg
Now calculate mass of air left,where me is the mass of air left.
me=m2-m1
me=0.715-0.357
mass of air left=me=0.357 Kg
To find heat transfer we need to apply energy balance equation.
Where me=m1-m2
And as the temperature remains constant,hence the enthalpy also remains constant.
h1=h2=he=h
Q=(me-(m1-m2))*h
me=m1-me
Thus heat transfer=Q=0
Answer:
Part A:
CPI cannot be negative so it is not possible to for program to run two times faster.
Part B:
CPI reduced by =80%
Part C:
New Execution Time=
Increase in speed=
Explanation:
FP Instructions=50*106=5300
INT Instructions=110*106=11660
L/S Instructions=80*106=8480
Branch Instructions=16*106=1696
Calculating Execution Time:
Execution Time=
Execution Time=
Execution Time=
Part A:
For Program to run two times faster,Execution Time (Calculated above) is reduced to half.
New Execution Time=
CPI cannot be negative so it is not possible to for program to run two times faster.
Part B:
For Program to run two times faster,Execution Time (Calculated above) is reduced to half.
New Execution Time=
CPI reduced by =80%
Part C:
New Execution Time=
New Execution Time=
Increase in speed=