Answer:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Explanation:
Using Boyle's law

Given ,
V₁ = 3.6 L
V₂ = ?
P₁ = 1.0 atm
P₂ = 13.3 atm (From correct source)
Using above equation as:




The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Answer: 0.548J/g°C
Explanation:
Q = s × m × DeltaT
Q = Heat (J)
S = Specific Heat Capacity
M = mass (g)
DeltaT = Change in temperature (°C)
0.158Kg x 1000 = 158g
2.510J = s x 158g x (61°C-32°C)
2.510J/(158g x 29°C) = s
S = 0.54779.... J/g°C
S = 0.548 J/g°C
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
Answer: False. Hope this helps!!
Answer:
no he just repeated the steps and made more of the same cleaner my guy
Explanation: