A) 3.0 atm
B) 1 atm
C) 3.33 atm
D) 2.4 atm
Correct Answer: D
Solution :
\[P{{'}_{{{N}_{2}}}}+P{{'}_{{{H}_{2}}O}}\,\,=\,\,1\,\,atm\] \[P{{'}_{{{H}_{2}}O}}\,\,=\,\,0.3\,\,atm\] \[P{{'}_{{{N}_{2}}}}=\,\,0.7\,atm\] Now new pressure of \[{{N}_{2}}\] in another vessel of volume V/3 at same temperature T is given by \[P{{'}_{{{N}_{2}}}}\times \frac{{{V}_{1}}}{3}=\,\,0.7\,atm\] \[\therefore \,\,\,\,\,\,\,=\,\,\,P{{''}_{{{N}_{2}}}}=\,\,21\,\,atm\] Since aqueous tension remains constant, and thus total pressure in new vessel \[P{{'}_{{{N}_{2}}}}+\,\,P{{'}_{{{H}_{2}}O}}\,\,= 2.1 + 0.3 = 2.4 atm~~~~\]You need to login to perform this action.
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