For a van der Waals gas X, the Boyle temperature (van der Waals temperature) is 1260 K. Calculate the van der Waals constant 'a' if 1 mole of the gas occupies 0.2 L at its critical pressure and critical volume. [Given: R = 1/12 L atm K⁻¹ mol⁻¹; answer in atm L² mol⁻²]

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1/3. V_c = 3b gives b = 0.2/3 L/mol. T_B = a/(Rb) = 1260 K, so a = R * b * 1260 = (1/12)*(0.2/3)*1260 = (1/12)*(84/3) = 7/3... Let me recalculate: a = 1260 * R * b = 1260 * (1/12) * (0.2/3) = 1260 * 0.2/36 = 252/36 = 7. Hmm, let me try b = Vc/3 more carefully. If Vc = 0.2, b = 0.2/3. a = 1260*R*b = 1260*(1/12)*(0.2/3) = 105*(0.2/3) = 21/3 = 7. That gives a=7, not matching options. Alternative: maybe Vc = 3b refers to molar critical volume, so b = 0.2/3 still. Try T_B = a/(Rb): 1260 = a/[(1/12)*b]. So a = 1260b/12 = 105b. With b=0.2/3: a = 105*0.2/3 = 21/3 = 7. Still 7. Perhaps the 0.2 L is not Vc but just the volume at Pc, Tc for real calculation using PcVc = (3/8)RTc. Pc*0.2 = (3/8)*(1/12)*

For a van der Waals gas X, the Boyle temperature (van der Waals temperature) is 1260 K. Calculate the van der Waals constant 'a' if 1 mole of the gas occupies 0.2 L at its critical pressure and critical volume. [Given: R = 1/12 L atm K⁻¹ mol⁻¹; answer in atm L² mol⁻²]

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