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Exercises · 5.4

Q.At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?

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Step 1 – Set up the density expression for each gas

From pV=nRTpV=nRT with n=m/Mn=m/M:  d=pMRT\ d = \dfrac{pM}{RT} (derived in Exercise 3).

At the SAME temperature (0°C) for both gases, RR and TT are common to both, so:

doxide=poxide MoxideRT,dN2=pN2 MN2RTd_{\text{oxide}} = \frac{p_{\text{oxide}}\,M_{\text{oxide}}}{RT}, \qquad d_{\text{N}_2} = \frac{p_{\text{N}_2}\,M_{\text{N}_2}}{RT}

Step 2 – Equate the densities (given equal)

doxide=dN2  ⟹  poxide MoxideRT=pN2 MN2RTd_{\text{oxide}} = d_{\text{N}_2} \implies \frac{p_{\text{oxide}}\,M_{\text{oxide}}}{RT} = \frac{p_{\text{N}_2}\,M_{\text{N}_2}}{RT}

The RTRT cancels:

poxide Moxide=pN2 MN2p_{\text{oxide}}\,M_{\text{oxide}} = p_{\text{N}_2}\,M_{\text{N}_2}

Step 3 – Substitute the data

poxide=2 bar,pN2=5 bar,MN2=28 g mol−1p_{\text{oxide}} = 2\ \text{bar}, \quad p_{\text{N}_2} = 5\ \text{bar}, \quad M_{\text{N}_2} = 28\ \text{g mol}^{-1} …

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