The figure below shows a thermodynamic process followed by 140 m g of helium. A
ID: 2251662 • Letter: T
Question
The figure below shows a thermodynamic process followed by 140mg of helium.
A Determine the pressure (in
atm) of the gas at points 1, 2, and 3.
B Determine the temperature (in
?C) of the gas at points 1, 2, and 3.
C Determine the volume (in
cm3) of the gas at points 1, 2, and 3.
D How much work is done on the gas during each of the three segments?
E How much heat energy is transferred to or from the gas during each of the three segments?
Explanation / Answer
Mass of helium m =140 mg
= 0.14 g
mass of 1 mole helium M = 2 g
So, No.of moles n = m / M
= 0.07 mol
temp at point 1 is T = 133 + 273 = 406K
volume at point a is V = 1000 cm ^ 3
= 10 ^ -3 m ^ 3
from gas law PV = nRT
from this pressure at point 1 is P = nRT / V
where R = gas constant = 8.314 J / mol K
from this we find P value
Pressure at point 2 is P ' = 5 P
volume at point 2 is V ' = V
At constant volume , ( P ' / P ) = ( T ' / T)
So, temp at point 2 is T ' = T P ' /P
temp at point 3 is T " = temp at point2 Since it is isothermal process
T " = T '
process 3 ---> 1 is isobaric process
Inthis process V " / V = T " / T
So, Volume at point 3 is V " = V T " /T
Pressure at point 3 is P " = P
Note : Temp in celsius scale = temp in kelvin scale -273
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