JEE Advanced levelSingle correct optionDifficulty 3 of 3rg054
The compressibility factor of a gas at 300K is found to be 0.9850 at 5.00atm and 0.9700 at 10.0atm. Assuming Z=1+B′P holds, the value of B′ and the second virial coefficient B=B′RT (R=0.0821LatmK−1mol−1) are
AB′=−3.00×10−3atm−1, B=−73.9mLmol−1
BB′=+3.00×10−3atm−1, B=+73.9mLmol−1
CB′=−1.50×10−3atm−1, B=−36.9mLmol−1
DB′=−3.00×10−3atm−1, B=−7.39mLmol−1
Answer: A
Slope B′=10.0−5.000.9700−0.9850=5.00−0.0150=−3.00×10−3atm−1 (both points are consistent with intercept 1). Then B=B′RT=−3.00×10−3×0.0821×300=−0.0739Lmol−1=−73.9mLmol−1.