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The volume 10 moles of an ideal gas at 10 atm and 500 K is

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82 L

41 L

20.5 L

$\frac{82}{3} L$

answer is B.

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Detailed Solution

n = 10 mole

P = 10 atm

T = 500 K

R = 0.0821 L-atm $K^{- 1} m o l^{- 1}$

PV = nRT

10 x V = 10 x 0.082 x 500

V = 41 L

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