An ideal gas at 0.80 atmospheres and 87oC occupies 0.450 liter. How many moles are in the sample? (R =0.0821 liter-atm/mole-K)
According to ideal gas law pV=νRTp V = \nu R TpV=νRT, from where the amount of substance is ν=pVRT=0.8atm⋅0.45l0.0821l⋅atmmol⋅K⋅360.15K≈0.012mol\nu = \frac{p V}{R T} = \frac{0.8 atm \cdot 0.45 l}{0.0821 \frac{l \cdot atm}{mol \cdot K} \cdot 360.15 K} \approx 0.012 molν=RTpV=0.0821mol⋅Kl⋅atm⋅360.15K0.8atm⋅0.45l≈0.012mol.
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