JEE Main levelNumerical answerDifficulty 2 of 3rg070
One mole of methane (a=2.25L2atmmol−2, b=0.0428Lmol−1) occupies 0.500L at a pressure of 40.0atm. Using the van der Waals equation and R=0.0821LatmK−1mol−1, the temperature in K (nearest integer) is
Numerical answer type, answer in K, tolerance 2.
Answer: 273 K
T=R(P+a/Vm2)(Vm−b)=0.0821(40.0+2.25/0.250)(0.500−0.0428)=0.082149.0×0.4572=0.082122.403=272.9K, i.e. 273K.