Answer:
M=28.88 gm/mol
Explanation:
Given that
T= 95 K
P= 1.6 atm
V= 4.87 L
m = 28.6 g
R=0.08206L atm .mol .K
We know that gas equation for ideal gas
P V = n R T
P=Pressure , V=Volume ,n=Moles,T= Temperature ,R=gas constant
Now by putting the values
P V = n R T
1.6 x 4.87 = n x 0.08206 x 95
n=0.99 moles
We know that number of moles given as
M=Molar mass
M=28.88 gm/mol
Answer:
um d. but I am guessing this ans
A. 320 g
B. 160 g
C. 80 g
D. 40 g
Answer:=14,160 kJ
Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below
Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase
C MPa m^3/kg kJ/kg kJ/kg
1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture
2 260 4.689 0.0422 2599 2797 1 Saturated Vapor
The mass initially contained in the tank is m1 = V/v1
m1 =0.85 m^3 /0.02993 m^3 /kg
= 28.4 kg
The mass finally contained in the tank is
m2 =V2/v
= 0.85 m^3 /0.0422 m^3 /kg
= 20.14 kg
The heat transfer is then
Qcv = m2u2 − m1u1 − he(m2 − m1)
Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ
