| (i) 1 L atm, |
| (ii) 1 erg, |
| (iii) 1 J |
| (iv) kcal, |
A) I = II = Ill = IV
B) I < II < III < IV
C) II < III < I < IV
D) IV < l < III < II
Correct Answer: C
Solution :
\[R=0.0821\,L\,atm\,mo{{l}^{-1}}{{K}^{-1}}\] \[=8.314\,\times {{10}^{7}}ergs\,mo{{l}^{-1}}{{K}^{-1}}\] \[=8.314\,Jmo{{l}^{-1}}{{K}^{-1}}\] \[=0.002\,kcal\,mo{{l}^{-1}}\,{{K}^{-1}}\] \[\therefore \]\[1\,L\,\,atm=\frac{R}{0.0821}mol\,K\] \[1\,erg=\frac{R}{8.314\times {{10}^{7}}}mol\,K\] \[1\,J=\frac{R}{8.314}\,mol\,K\] \[1\,kcal=\frac{R}{0.002}\,mol\,K\] Thus, II < III < I < IV
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