Answer:
c) 1.75 g/cm³
Explanation:
Given that
Radii of the A ion, r(c) = 0.137 nm
Radii of the X ion, r(a) = 0.241 nm
Atomic weight of the A ion, A(c) = 22.7 g/mol
Atomic weight of the X ion, A(a) = 91.4 g/mol
Avogadro's number, N = 6.02*10^23 per mol
Solution is attached below
Answer:
a) 22.5number
b) 22.22 m length
Explanation:
Given data:
Bridge length = 500 m
width of bridge = 12 m
Maximum temperature = 40 degree C
minimum temperature = - 35 degree C
Maximum expansion can be determined as
where , \alpha is expansion coefficient degree C
SO,
number of minimum expansion joints is calculated as
b) length of each bridge
Answer:
a) 358.8K
b) 181.1 kJ/kg.K
c) 0.0068 kJ/kg.K
Explanation:
Given:
P1 = 100kPa
P2= 800kPa
T1 = 22°C = 22+273 = 295K
q_out = 120 kJ/kg
∆S_air = 0.40 kJ/kg.k
T2 =??
a) Using the formula for change in entropy of air, we have:
∆S_air =
Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K
Solving, we have:
[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]
Solving for T2 we have:
Taking the exponential on the equation (both sides), we have:
[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]
b) Work input to compressor:
= 184.1 kJ/kg
c) Entropy genered during this process, we use the expression;
Egen = ∆Eair + ∆Es
Where; Egen = generated entropy
∆Eair = Entropy change of air in compressor
∆Es = Entropy change in surrounding.
We need to first find ∆Es, since it is unknown.
Therefore ∆Es =
∆Es = 0.4068kJ/kg.k
Hence, entropy generated, Egen will be calculated as:
= -0.40 kJ/kg.K + 0.40608kJ/kg.K
= 0.0068kJ/kg.k