The half-life of krypton-91 (91Kr) is 10 s. At time t = 0 a heavy canister contains 7 g of this radioactive gas. (a) Find a func
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Answer:
ΔH =
Explanation:
In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.
The heat transfer is represented by
=
where
= the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.
= the heat of combustion
Also, we know that the total heat change of the any system is
ΔH = ΔQ + ΔW
where
ΔH = the total heat absorbed by the system
ΔQ = the internal heat absorbed by the system which in this case is
ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0
substituting into the heat change equation
ΔH = + 0
==> ΔH =
Answer:
a. 478.69 K
b. 939.43
c. 19.30 J
d. 64.5J
Explanation:
From the question, we can identify the following;
= 785
= 0.000785
= 400K
= 125 Kpa = 125 000 Pa
Using the ideal gas equation,
PV = nRT
where R is the molar gas constant = 8.314 ⋅Pa⋅
⋅
Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol
a. Steam temperature in K
To calculate this, we use the constant pressure process;
q = nΔH
Where q is 83.8J according to the question
Thus;
83.8 = 0.03 × [34980 + 35.5 - (34980 + 35.5
)]
83.8 = (0.03 × 35.5) ( - 400K)
83.8 = 1.065 ( - 400)
78.69 = ( - 400)
= 400 + 78.69
= 478.69 K
b. Final cylinder volume
To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)
/
=
/
=
/
= (785 × 478.69)/400
= 939.43
c. Work done by the system
Mathematically, the work done by the system is calculated as follows;
w = P(-
) = 125 KPa ( 939.43 - 785) = 19.30 J
d. Change in internal energy of the steam in J
ΔU = q - w = 83.8 - 19.3 = 64.5J