10 moles of ideal gas confined to a volume of 10 L is released into atmosphere at 300 K where the pressure is 1 bar. The work done by the gas is (R = 0.083 L-bar k-1 mol-1).

Question

10 moles of ideal gas confined to a volume of 10 L is released into atmosphere at 300 K where the pressure is 1 bar. The work done by the gas is (R = 0.083 L-bar k-1 mol-1). (A) 249 L-bar (B) 259 L-bar (C) 239 L-bar (D) 220 L-bar

Tardigrade Admin · Accepted Answer

Initial volume = 10 L Final volume=(nRT/P)=(10×0.083×300/1)=249 L ∴ Δ V=249-10=239 L ∴ Wpv=-Pext. Δ V=1×239=239 L-bar

Q. $10$ moles of ideal gas confined to a volume of $10 L$ is released into atmosphere at $300 K$ where the pressure is $1$ bar. The work done by the gas is ( $R = 0.083 L - ba r k^{- 1} m o l^{- 1}$ ).

$249 L - ba r$

$259 L - ba r$

$239 L - ba r$

$220 L - ba r$

Initial volume $= 10 L$
Final volume $= \frac{n RT}{P} = \frac{10 \times 0.083 \times 300}{1} = 249 L$
$∴ Δ V = 249 - 10 = 239 L$
$∴ W_{p v} = - P_{e x t} .Δ V = 1 \times 239 = 239 L - ba r$

Q. 10 moles of ideal gas confined to a volume of 10L is released into atmosphere at 300 K where the pressure is 1 bar. The work done by the gas is (R=0.083L−bark−1mol−1).

249L−bar

259L−bar

239L−bar

220L−bar