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NCERT Class XI Chemistry States of Matter Solutions

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Question : 12 of 23
Marks: +1, -0
Calculate the temperature of 4.0 mol of a gas occupying 5 dm3\mathrm{dm}^3 at 3.32 bar. (R = 0.083 bar dm3K−1mol−1\mathrm{dm}^3 \mathrm{K}^{-1} \mathrm{mol}^{-1})
Solution:  
n = 4 moles, P = 3.32 bar, V = 5 dm3\mathrm{dm}^3, R = 0.083 bar dm3K−1mol−1\mathrm{dm}^3 \mathrm{K}^{-1} \mathrm{mol}^{-1}
We know that PV = nRT
T = PVnR\frac{PV}{nR} = 3.32×54×0.083\frac{3.32 \times 5}{4 \times 0.083} ⇒ T = 16.60.332\frac{16.6}{0.332} = 50K
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