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I surfed through the internet and visited a number of chemistry websites but only in vain. I couldn't find an answer.
Please help! Your quick response will be appreciated.

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$1\ \mathrm{dm^3} = 1\times 10^{-3}\ \mathrm{m}^3 $

$1\ \mathrm{atm = 1013.25\ mbar = 101.325 kPa = 101325\ {kg}\cdot{}m^{-1}\cdot s^{-2}}$

$1\ \mathrm{atm \cdot dm^3} = 1\times 10^{-3}\cdot 101325\ \mathrm{m^3 \cdot kg \cdot{}m^{-1}\cdot s^{-2}} = 101.325\ \mathrm{kg \cdot{}m^{2}\cdot s^{-2}}$

$ 1\ \mathrm{atm \cdot dm^3 = 101.325\ J = 101.325\cdot 0.2390\ cal= 24.217\ cal}$

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General Gas Constant, $\mathcal{R} = 0.0821\:\mathrm{dm^3\cdot atm\cdot K^{-1} mol^{-1}}$
General Gas Constant, $\mathcal{R} = 1.989\:\mathrm{cal\cdot K^{-1} mol^{-1}}$

\begin{aligned} \mathcal{R} &: \mathcal{R}\\ 0.0821 \:\mathrm{dm^3\cdot atm\cdot K^{-1} mol^{-1}} &: 1.989\:\mathrm{cal\cdot K^{-1} mol^{-1}} \\ 0.0821\:\mathrm{dm^3 atm} &: 1.989 \:\mathrm{cal} \\ 1\:\mathrm{dm^3 atm} &: (1.989\:\mathrm{cal}\cdot 1\:\mathrm{dm^3 atm})/ 0.0821\:\mathrm{dm^3 atm}$ \\ 1\:\mathrm{dm^3 atm} &: 24.22\:\mathrm{cal}\\ \end{aligned}

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