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JEE Main levelNumerical answerDifficulty 2 of 3rg213

A gas has $B^{'} = - 1.80 \times 1 0^{- 3} at m^{- 1}$ at a certain temperature. Using $Z = 1 + B^{'} P$ , its compressibility factor at $15.0 atm$ (three decimals) is

Numerical answer type, tolerance 0.002.

Answer: 0.973

Z = 1 + B^{'} P = 1 + (- 1.80 \times 1 0^{- 3}) (15.0) = 1 - 0.0270 = 0.9730

, i.e.

0.973

.

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