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4.0g of argon has pressure P at temperature T K in a vessel . On keeping the sample at $50^{\large\circ}C$ higher temperature 0.8g gas was given out to maintain the pressure P. The original temperature was:


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Initially for Argon $P\times V = \large\frac{4}{40}\times R\times T$
(molecular weight of Ar = 40)
On heating,
$P\times V = \large\frac{3.2}{40} \times R\times (T+50)$
$\therefore 4T = 3.2T + 160$
$T = \large\frac{160}{0.8}$
Hence answer is (c)
answered Mar 5, 2014 by sharmaaparna1

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