A) 25.215 atm
B) 31.205 atm
C) 45.215 atm
D) 15.210 atm
Correct Answer: A
Solution :
\[^{n}{{O}_{2}}=\frac{4}{32}=\frac{1}{8}mol\] \[^{n}{{H}_{2}}=\frac{2}{2}=1\,\,mol\] \[\operatorname{n}={{\,}^{n}}{{O}_{2}}+{{\,}^{n}}{{H}_{2}}=\frac{1}{8}+1=\frac{9}{8}\,\,mol\] [ideal gas eqn] \[\operatorname{p}=\frac{nRT}{v}\] \[=\frac{\frac{9}{8}\times 0.0821\times 273}{1}=25.215\,\,atm\]You need to login to perform this action.
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