The pressure inside the can upon cooling is 0.4 atm.
<u>Explanation:</u>
Given -
Initial Temperature, T1 = 908°C = 908 + 273 K = 1181 K
Final Temperature, T2 = 208°C = 208 + 273 K = 481 K
Pressure upon cooling, P2 = ?
Using Gay Lussac's law:
P1/T1 = P2/T2
P2 = P1 X T2 / T1
P2 = 1 atm X 481 / 1181
P2 = 0.4 atm
Therefore, the pressure inside the can upon cooling is 0.4 atm.
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Answer:
The time necessary to purge 95% of the NaOH is 0.38 h
Explanation:
Given:
vfpure water(i) = 3 m³/h
vNaOH = 4 m³
xNaOH = 0.2
vfpure water(f) = 2 m³/h
pwater = 1000 kg/m³
pNaOH = 1220 kg/m³
The mass flow rate of the water is = 3 * 1000 = 3000 kg/h
The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg
When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg
The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³
The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h
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