A 29.1 liter tank contains ideal helium gas at 35.8°C and a pressure of 21.8 atm How many moles of gas are in the tank?
1 answer:
Answer:
No. of moles, n = 25.022 moles
Given:
Volume of gas in tank, V = 29.1 l
Temperate of gas, T = = 273 + 35.8 = 308.8 K
Pressure of gas, P = 21.8 atm
Solution:
Making use of the ideal gas equation which given as:
PV = nRT
where
R = Rydberg's constant = 0.0821 L-atm/mol-K
Re-arranging the above formula for 'n' and putting the values in the above formula:
n = 25.022
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Answer:
(a) The resistor disspates 103680 joules during a 24-hour period.
(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.
(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.
Explanation:
(a) The amount of heat dissipated (), measured in joules, by the cylindrical resistor is the power multiplied by operation time (
), measured in hours. That is:
(1)
If we know that and
, then the amount of heat dissipated by the resistor is:
The resistor disspates 103680 joules during a 24-hour period.
(b) The heat flux (), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder (
), measured in square meters:
(2)
(3)
Where:
- Diameter, measured in meters.
- Length, measured in meters.
If we know that ,
and
, the heat flux of the resistor is:
The heat flux of the resistor is approximately 4340.589 watts per square meter.
(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces (), no unit, is the ratio of the top and bottom surfaces to total surface:
(3)
If we know that and
, then the fraction is:
The fraction of heat dissipated from the top and bottom surfaces is 0.045.