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IV. Exercises · Q2

Q.Consider a mixture of 2 mol of helium and 4 mol of oxygen. Compute the speed of sound in this gas mixture at 300 K.

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Step 1. For a gas mixture, the effective specific heats are mole-weighted averages: for helium (monatomic, CV1=32RC_{V1}=\tfrac32R) and oxygen (diatomic, CV2=52RC_{V2}=\tfrac52R), with n1=2 moln_1=2\ \text{mol} He and n2=4 moln_2=4\ \text{mol} O2_2: CV,mix=n1CV1+n2CV2n1+n2=2(1.5R)+4(2.5R)6=3R+10R6=13R6C_{V,mix}=\dfrac{n_1C_{V1}+n_2C_{V2}}{n_1+n_2}=\dfrac{2(1.5R)+4(2.5R)}{6}=\dfrac{3R+10R}{6}=\dfrac{13R}{6}.

Step 2. Correspondingly CP,mix=CV,mix+R=13R6+R=19R6C_{P,mix}=C_{V,mix}+R=\dfrac{13R}{6}+R=\dfrac{19R}{6}, so γmix=CP,mix/CV,mix=19/13≈1.4615\gamma_{mix}=C_{P,mix}/C_{V,mix}=19/13\approx1.4615.

Step 3. The mean molar mass is Mmix=n1M1+n2M2n1+n2=2(4)+4(32)6=8+1286=1366≈22.67 g/mol=0.02267 kg/molM_{mix}=\dfrac{n_1M_1+n_2M_2}{n_1+n_2}=\dfrac{2(4)+4(32)}{6}=\dfrac{8+128}{6}=\dfrac{136}{6}\approx22.67\ \text{g/mol}=0.02267\ \text{kg/mol}.

Step 4. Speed of sound: v=γmixRTMmix=1.4615×8.314×3000.02267=3645.60.02267=160834≈401.0 m/sv=\sqrt{\dfrac{\gamma_{mix}RT}{M_{mix}}}=\sqrt{\dfrac{1.4615\times8.314\times300}{0.02267}}=\sqrt{\dfrac{3645.6}{0.02267}}=\sqrt{160834}\approx401.0\ \text{m/s}.

✓Final answer

v≈400.9 m/sv \approx 400.9\ \text{m/s}

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