A) 0.6816 atm
B) 0.6618 atm
C) 0.8616 atm
D) zero
Correct Answer: A
Solution :
Given that, Reduced volume, \[\phi =10.2\] Reduced temperature, \[\theta =0.7\] Critical pressure, \[{{p}_{c}}=4.25\] atm, p = ? According to reduced equation of state \[\left( \pi +\frac{3}{{{\phi }^{2}}} \right)\,(3\phi -1)=8\theta \] \[\left( \pi +\frac{3}{10.2\times 10.2} \right)\,(3\times 10.2-1)=8\times 0.7\] \[\left( \pi +\frac{3}{104.04} \right)\,(30.6-1)=5.6\] \[(\pi +0.0288)\,(29.6)\,=5.6\] \[\pi +0.0288=\frac{5.6}{29.6}=0.1892\] \[\pi =0.1892-0.0288\] = 0.1604 Now, we know that \[\pi =\frac{p}{{{p}_{c}}}\] \[\therefore \] \[p=\pi \times {{p}_{c}}\] \[p=0.1604\times 4.25\] = 0.6816 atm
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