For one gram mole of a gas,the value of R in | vedclass

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(A) For $1$ gram mole of a gas,the ideal gas equation is $PV = RT$.The universal gas constant $R$ is approximately $8.314 \; J/(mol \cdot K)$.To conve

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$2$

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(A) For $1$ gram mole of a gas,the ideal gas equation is $PV = RT$.The universal gas constant $R$ is approximately $8.314 \; J/(mol \cdot K)$.To convert this value into calories,we use the conversion factor $1 \; cal \approx 4.184 \; J$ (often approximated as $4.2 \; J$ or $4.18 \; J$ in physics problems).$R = \frac{8.314 \; J/K}{4.184 \; J/cal} \approx 1.987 \; cal/(mol \cdot K)$.Rounding this value to the nearest whole number,we get $R \approx 2 \; cal/(mol \cdot K)$.Therefore,the correct option is $(A)$.

For one gram mole of a gas,the value of $R$ in the equation $PV = RT$ is nearly ...... $cal/K$

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