For one mole of an ideal gas, the slope of $V$ versus $T$ curve at a constant pressure of $2 a t m$ is $X L m o l^{- 1} K^{- 1}$ . The value of the ideal universal gas constant $R$ in terms of $X$ is:

X L a t m m o l^{- 1} K^{- 1}

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\frac{X}{2} L a t m m o l^{- 1} K^{- 1}

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2 X L a t m m o l^{- 1} K^{- 1}

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2 X a t m L^{- 1} m o l^{- 1} K^{- 1}

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Solution

The correct option is C $2 X L a t m m o l^{- 1} K^{- 1}$
At constant pressure and same number of moles, Volume of the gas is the directly proportional to the temperature of the gas. The slope of V versus T graph i.e slope= $\frac{Δ V}{Δ T}$ = $X$ $L {m o l}^{- 1} K^{- 1}$
But, $\frac{Δ V}{Δ T}$ = $\frac{R}{P}$ (no. of moles = 1)
Now, $P = 2$
Therefore, $R$ = $2 X$ $L atm$ ${m o l}^{- 1} K^{- 1}$

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