The value of gas constant R when pressure is expressed in $dyn m^{- 2}$ and volume in ${mm}^{3}$ is

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$8.314 \times 10^{16} ({dynm}^{- 2}) ({mm}^{3}) K^{- 1} {mol}^{- 1}$

$8.314 \times 10^{10} (dyn m^{- 2}) ({mm}^{3}) K^{- 1} {mol}^{- 1}$

$8.314 \times 10^{14} (dyn m^{- 2}) ({mm}^{3}) K^{- 1} {mol}^{- 1}$

$8.314 \times 10^{12} ({dynm}^{- 2}) ({mm}^{3}) K^{- 1} {mol}^{- 1}$

answer is C.

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Detailed Solution

$1 {dynm}^{- 2} = (1 {gcms}^{- 2}) {(10^{2} cm)}^{- 2} = 10^{- 4} g {cm}^{- 1} s^{- 2}$

$\begin{array}{l} 1 atm = (76.0 cm) (13.595 g {cm}^{- 3}) (980.6 {cms}^{- 2}) = 1.013 \times 10^{6} g {cm}^{- 1} s^{- 2} = 1.013 \times 10^{10} {dynm}^{- 2} \\ V_{m} = 22.414 {dm}^{3} {mol}^{- 1} = 22.414 \times 10^{6} {mm}^{3} {mol}^{- 1} \\ R = \frac{p V}{T} = \frac{(1.103 \times 10^{10} {dynm}^{- 2}) (22.414 \times 10^{6} {mm}^{3} {mol}^{- 1})}{(273 K)} = 8.314 \times 10^{14} ({dynm}^{- 2}) ({mm}^{3}) K^{- 1} {mol}^{- 1} \end{array}$