Number of particles inside a sugar bowl = 4.4 x 10²³
<h3>Further explanation</h3>Given
The molar mass of sugar = 342.3 g/mol
mass of sugar = 250 g
Required
Number of particles
Solution
1 mol = 6.02 x 10²³ particles
mol of sugar :
mol = 250 g : 342.3 g/mol = 0.73 mol
Number of particles :
N = n x No
N = 0.73 x 6.02 x 10²³
N = 4.39 x 10²³ ≈ 4.4 x 10²³
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