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Exercise · Q20

Q.Calculate the volume occupied by 2 mol2\ \text{mol} of an ideal gas at 300 K300\ \text{K} and 1 atm1\ \text{atm} pressure. (R=0.0821 L atm K−1mol−1R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1})

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Rearranging the ideal gas equation PV=nRTPV = nRT for volume: V=nRTPV = \dfrac{nRT}{P}. Substituting n=2 moln = 2\ \text{mol}, R=0.0821 L atm K−1mol−1R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}, T=300 KT = 300\ \text{K}, and P=1 atmP = 1\ \text{atm}: V=2×0.0821×3001=49.261=49.26 LV = \frac{2\times0.0821\times300}{1} = \frac{49.26}{1} = 49.26\ \text{L} …

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