The work for the process of nitrogen compression where the external pressure is constant at 1.00 atm, the initial temperature is 200 K, and the final temperature is 64 K is 3719.3 J.
The work for this compression process is given by:
[tex] W = -p(\Delta V) = -p(V_{f} - V_{i}) [/tex] (1)
Where:
p: is the pressure = 1.00 atm
[tex] V_{f}[/tex]: is the final volume
[tex]V_{i}[/tex]: is the initial volume
We can find the initial and final volume with the Ideal Gas Law:
[tex] pV = nRT [/tex] (2)
Where:
n: is the number of moles
R: is the gas constant = 0.08206 L*atm/(K*mol)
T: is the temperature
After solving equation (2) for V and entering into equation (1), we have:
[tex] W = -p(\frac{nRT_{f}}{p_{f}} - \frac{nRT_{i}}{p_{i}}) [/tex]
Since [tex]p_{f} = p_{i} = p [/tex], we have:
[tex] W = -nR(T_{f} - T_{i}) [/tex] (3)
Where:
[tex]T_{f}[/tex]: is the final temperature = 64 K
[tex]T_{i}[/tex]: is the initial temperature = 200 K
First, we need to find the number of moles of nitrogen
[tex] n = \frac{m}{M} [/tex] (4)
Where:
m: is the mass of nitrogen = 92 g
M: is the molecular weight of nitrogen = 28 g/mol
The number of moles is (eq 4):
[tex] n = \frac{m}{M} = \frac{92 g}{28 g/mol} = 3.29 \:moles [/tex]
Now we can find the work (eq 3)
[tex] W = -nR(T_{f} - T_{i}) = -3.29 \:moles*0.08206 L*atm/(K*mol)(64 K - 200 K) = 36.716 L*atm*\frac{101.3 J}{1 L*atm} = 3719.3 J [/tex]
Therefore, the work for the compression process is 3719.3 J.
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