A) 272.84 kcal
B) 260 kcal
C) 180 kcal
D) 130 kcal
Correct Answer: A
Solution :
At STP \[16\,g\,{{O}_{2}}\] or \[1/2\,mol\,{{O}_{2}}\] will occupy \[11.2L\]. If volume is doubed i.e., \[({{v}_{2}}-{{v}_{1}})\] \[=22.4-11.2=11.2\] \[\therefore \] \[W=p({{V}_{2}}-{{V}_{1}})\] \[=1(11.2)=11.2\,L-atm\] \[=\frac{11.2\times 2}{0.0821}\,\text{cal}\] \[=272.84\,\,cal\]
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