A 0.50 mol sample of an ideal gas is kept at 273 degC and 2.0 atm. Taking R = 0.0821 L atm K^-1 mol^-1, its volume in litres (two decimals) is

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A $0.50 mol$ sample of an ideal gas is kept at $27 3^{\circ} C$ and $2.0 atm$ . Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , its volume in litres (two decimals) is

Numerical answer type, answer in L, tolerance 0.02.

T = 273 + 273 = 546 K

V = \frac{n RT}{P} = \frac{0.50 \times 0.0821 \times 546}{2.0} = 11.2067 \approx 11.21 L