Answer:
The specific gravity of the unkown liquid is 15.
Explanation:
Gauge pressure, at the bottom of the tank in this case, can be calculated from
where and
are the height of the column of oil and the unkown liquid, respectively. Writing for
, we have
Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.
Answer:
(a) ΔU=747J
(b) γ=1.3
Explanation:
For (a) change in internal energy
According to first law of thermodynamics the change in internal energy is given as
ΔU=Q-W
Substitute the given values
ΔU=970J-223J
ΔU=747J
For(b) γ for the gas.
We can calculate γ by ratio of heat capacities of the gas
γ=Cp/Cv
Where Cp is the molar heat capacity at constant pressure
Cv is the molar heat capacity at constant volume
To calculate γ we first need to find Cp and Cv
So
For Cp
As we know
Q=nCpΔT
Cp=(Q/nΔT)
From relation of Cv and Cp we know that
Cp=Cv+R
Where R is gas constant equals to 8.314J/mol.K
So
So
γ=Cp/Cv
γ=[(37J/mol.K) / (28.687J/mol.K)]
γ=1.3