A) 4.05 atm
B) 2.025 atm
C) 3.84 atm
D) 1.92 atm
Correct Answer: D
Solution :
[d] \[{{N}_{2}}\] \[\to \] 2N At t=0 \[\frac{1.4}{28}=\frac{1}{20}\] 0 At t=\[{{t}_{f}}\] \[\frac{1}{20}-x\] \[2x\] But, \[x=30\]% of \[\frac{1}{20}=\frac{3}{200}\] Final number of mole \[\frac{1}{20}-x+2x=\frac{1}{20}+x=\frac{1}{20}+\frac{3}{200}=\frac{13}{200}\] \[\therefore P=\frac{13}{200}\times \frac{0.2821\times 1800}{5}=1.92atm\]
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