A) 0.082 L-atm \[{{K}^{-1}}\,mo{{l}^{-1}}\]
B) 0.082 \[L-at{{m}^{-1}}\] K mol
C) 0.082 L-atm K mol
D) 0.082 \[L-at{{m}^{-1}}\] K mol
Correct Answer: A
Solution :
Key Idea We know \[pV=nRT\] \[R=\frac{pV}{nT}\] \[R=\frac{pV}{nT}\] \[=\frac{pressure\times volume}{mole\times temperature}\] \[=\frac{work\text{ }or\text{ }energy}{mole\times temperature}\] R is expressed in the unit of work or energy\[mo{{l}^{-1}}\,{{K}^{-1}}\]. The different values of R are summarised below R = 0.0821 L atm \[mo{{l}^{-1}}\,{{K}^{-1}}\] \[=8.3143\times {{10}^{7}}\,erg\,\,mo{{l}^{-1}}\,{{K}^{-1}}\] \[=8.3143\,\,J\,\,mo{{l}^{-1}}\,{{K}^{-1}}\] \[=8.3143\,\,Nm\,\,mo{{l}^{-1}}\,{{K}^{-1}}\] \[=8.3143\times {{10}^{-3}}\,\,kJ\,\,mo{{l}^{-1}}\,{{K}^{-1}}\] =1.987 cal \[mo{{l}^{-1}}\,{{K}^{-1}}\] \[=5.189\times {{10}^{19}}\,\,eV\,\,mo{{l}^{-1}}\,{{K}^{-1}}\]
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