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Answer:
(a) The heat transferred is 2552.64 kJ
(b) The entropy change of the mixture is 1066.0279 J/K
Explanation:
Here we have
Molar mass of H₂ = 2.01588 g/mol
Molar mass of N₂ = 28.0134 g/mol
Number of moles of H₂ = 500/2.01588 = 248 moles
Number of moles of N₂ = 1200/28.0134 = 42.8 moles
P·V = n·R·T
V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³
Since the volume is doubled then
V₂ = 2 × 7.25 = 14.51 m³
At constant pressure, the temperature is doubled, therefore
T₂ = 600 K
If we assume constant specific heat at the average temperature, we have
Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂
cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K
cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K
Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ
b)
Where:
and
are the mole fractions of Hydrogen and nitrogen respectively.
Therefore, = 248 /(248 + 42.8) = 0.83
= 42.8/(248 + 42.8) = 0.1472
∴ = 1066.0279 J/K
Answer:
Explanation:
To find the expression for the position x of the particle as a function of time t you need to integrate the expression for acceleration two times. To find the integral constants you are already given the initial conditions for both the position and the velocity which are:
The first integration gives you the velocity:
You get the constant by using the second initial condition:
To get the expression for the position you need to integrate again:
where using the first initial condition.
You can now plot the graph. To get the total distance travelled you can integrate the expression for the velocity twice and sum the two integrals (from 0 to 1 and from 1 to 3):
20 m
