Answer:
1.95
Explanation:
Just did the test and it was correct!
Answer:
Explanation:
Given that:
the temperature 
= 250 °C= ( 250+ 273.15 ) K = 523.15 K
Pressure = 1800 kPa
a)
The truncated viral equation is expressed as:
where; B = - 
C = -5800 
R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹
Plugging all our values; we have
Multiplying through with V² ; we have
V = 2250.06 cm³ mol⁻¹
Z = 
Z = 
Z = 0.931
b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].
The generalized Pitzer correlation is :
The compressibility is calculated as:
Z = 0.9386
V = 2268.01 cm³ mol⁻¹
c) From the steam tables (App. E).
At 
V = 0.1249 m³/ kg
M (molecular weight) = 18.015 gm/mol
V = 0.1249 × 10³ × 18.015
V = 2250.07 cm³/mol⁻¹
R = 729.77 J/kg.K
Z =
Z = 
Z = 0.588
<span>Let's </span>assume that water vapor has ideal gas
behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>
</span><span>Where, P is the pressure of the gas (Pa), V
is the volume of the gas (m³), n is the number
of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 1 atm = 101325 Pa (standard pressure)
V = 13.97 L = 13.97 x 10</span>⁻³ m³<span>
n = ?
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 0 °C = 273 K (standard temperature)
<span>
By substitution,
</span>101325 Pa x 13.97x 10</span>⁻³
m³ = n x 8.314 J mol⁻¹ K⁻¹ x 273 K<span>
n = 0.624 mol
<span>
Hence, the moles of water vapor at STP is 0.624 mol.
According to the </span></span>Avogadro's constant, 1 mole of substance has 6.022 × 10²³ particles.
<span>
Hence, number of atoms in water vapor = 0.624 mol x </span>6.022 × 10²³ mol⁻¹
<span> = 3.758 x 10</span>²³<span>
</span>
Answer:
both are done due to the enviroment??
