Answer:
This solution is quite lengthy
Total system = nRT
n was solved to be 0.02575
nH20 = 0.2x0.02575
= 0.00515
Nair = 0.0206
PH20 = 0.19999
Pair = 1-0.19999
= 0.80001
At 15⁰c
Pair = 0.4786atm
I used antoine's equation to get pressure
The pressure = 0.50
2. Moles of water vapor = 0.0007084
Moles of condensed water = 0.0044416
Grams of condensed water = 0.07994
Please refer to attachment. All solution is in there.
Answer:
B) Iron (c=0.45 J/g°C)
Explanation:
Given that:-
Heat gain by water = Heat lost by metal
Thus,
Where, negative sign signifies heat loss
Or,
For water:
Mass = 120 g
Initial temperature = 21.8 °C
Final temperature = 24.5 °C
Specific heat of water = 4.184 J/g°C
For metal:
Mass = 40.2 g
Initial temperature = 99.3 °C
Final temperature = 24.5 °C
Specific heat of metal = ?
So,
<u>This value corresponds to iron. Thus answer is B.</u>
Answer:
The maximum pressure is 612.2 Pa
Explanation:
The pressure of the ice (P1) = 624 Pa
The temperature of the ice = 273.16 K
The maximum temperature the specimen = - 5 oC
= -5 + 273 = 268K
The maximum Pressure the freeze drying can be will be (P2) = ?
Using Pressure law, which shows the relationship between pressure and temperature.
P1 / T1 = P2 / T2
P2 T1 = P1 T2
P2 = P1 T2 / T1
P2 = 624 × 268 / 273.16
P2 = 612.2 Pa
The maximum pressure at which drying can be carried out is 612.2 Pa
Check the attached document more explanation. jjjjggggg
