| I. 1 L atm, |
| II. 1 erg, |
| III. 1 J |
| IV. kcal, |
A) I = II = III = IV
B) l < II < III < IV
C) ll < lll < 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 \[mo{{l}^{-1}}{{K}^{-1}}\] \[=83.314\,J\,mo{{l}^{-1}}{{K}^{-1}}\] \[=0.002\,kcal\,\,mo{{l}^{-1}}{{K}^{-1}}\] \[\therefore \] \[1\,L\,atm=\frac{R}{0.0821}molK\] \[1\,erg=\frac{R}{8.314\times {{10}^{7}}}molK\] \[1\,J=\frac{R}{8.314}mol\,K\] \[1\,kcal=\frac{R}{0.002}mol\,K\] Thus, II < III < I < IV
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