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Q.

The pressure exerted by 2.0 mol of a van der Waal gas (a = 366 k Pa dm6mol–2 and b = 0.05 dm3mol–1) at T= 300 K and V=12.1dm3 is (use R = 8.32 kPa dm3 K–1 mol–1)

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a

395.0kPa

b

406.0kPa

c

420.1kPa

d

435.6kPa

answer is B.

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Detailed Solution

P+an2V2(V-nb)=nRTSubstituting the given values and simplifying(P+9.99)(12)=2×8.314×300J=4988.4P = 406.0kPa
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