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neet CHEMISTRYMedium

A 10.0 L flask contains 64 g of oxygen at 27°C. (Assume O2O_2 gas is behaving ideally). The pressure inside the flask in bar is (Given R=0.0831R = 0.0831 L bar K−1mol−1K^{-1} mol^{-1})

A

2.5

B

498.6

C

49.8

D

4.9

Step-by-Step Solution

We know for ideal gas PV=nRTPV = nRT. Given V=10.0V = 10.0 L, mass of O2=64O_2 = 64 g, molar mass of O2=32O_2 = 32 g/mol, so n=64/32=2n = 64/32 = 2 mol. T=27+273=300T = 27 + 273 = 300 K. R=0.0831R = 0.0831 L bar K−1mol−1K^{-1} mol^{-1}. P=(nRT)/V=(2×0.0831×300)/10=4.986P = (nRT)/V = (2 \times 0.0831 \times 300) / 10 = 4.986 bar. The closest value is 4.9 bar.

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