Question #a3859

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Question #a3859

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Answer 1 · 2017-03-17T01:23:17

$T_{2} = 846 K$ $T_2 = "846 K"$ or $573^{\circ} C$ $573^@"C"$

Explanation:

From the Ideal Gas Laws we know that ${(\frac{P V}{T})}_{1} = {(\frac{P V}{T})}_{2}$ $((PV)/T)_1 = ((PV)/T)_2$

At constant pressure this is just ${(\frac{V}{T})}_{1} = {(\frac{V}{T})}_{2}$ $((V)/T)_1 = ((V)/T)_2$
Inserting our given values into this equation with the correct temperature values we get:

$\frac{0.300 L}{423 K} = \frac{0.600 L}{T_{2}}$ $"0.300 L"/"423 K" = "0.600 L"/T_2$ ; $T_{2} = \frac{0.600}{\frac{0.300}{423 K}}$ $T_2 = 0.600/(0.300/"423 K")$

$T_{2} = 846 K$ $T_2 = "846 K"$ or $573^{\circ} C$ $573^@"C"$

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Question #a3859

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