The figure(Figure 1) shows a thermodynamic process followed by 120 of helium. Pa

ID: 1914646 • Letter: T

Question

The figure(Figure 1) shows a thermodynamic process followed by 120 of helium. Part A Determine the pressure (inatm ) of the gas at points 1, 2, and 3. Part B Determine the temperature (inC) of the gas at points 1, 2, and 3. Part C Determine the volume (in cm^3) of the gas at points 1, 2, and 3. Part D How much work is done on the gas during each of the three segments? Part E How much heat energy is transferred to or from the gas during each of the three segments?

Explanation / Answer

Given :
Mass of helium m =120 mg = 0.12 g
mass of 1 mole helium M = 2 g
So, No.of moles n = m / M = 0.06 mol

Temp at point 1 is T = 133 + 273 = 406K
volume at point a is V = 1000 cm3= 10-3 m3
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 = 0.06*8.314*406/10-3 = 202529.04 Pascals

Pressure at point 2 is P' = 5 P = 1012645.2 Pascals
volume at point 2 is V' = V = 10-3 m3At constant volume , ( P ' / P ) = ( T ' / T)
So, temp at point 2 is T ' = T P ' /P = 5T = 2030 K
temp at point 3 is T " = temp at point2 Since it is isothermal process
T " = T ' = 2030 K

process 3 ---> 1 is isobaric process
Inthis process V " / V = T " / T
So, Volume at point 3 is V " = V T " /T = 5V = 5*10-3 m3
Pressure at point 3 is P " = P = 202529.04 Pascals

Work done = area occupied by cycle = 1629.7896 - 810.1162 = 819.6734 J

Now by 1st law of thermod we have: U = Q-W
And here in cyclic process U = 0. Hence Q = W = 819.6734 J

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The figure(Figure 1) shows a thermodynamic process followed by 120 of helium. Pa

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