Question #c6627

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Question #c6627

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Answer 1 · 2014-12-03T14:29:04

The final temperature is $T_{2}$ $T_2$ =806.9 K

Using the ideal gas law, $P V = n R T$ $PV=nRT$ , and assuming the volume is being held constant, you will have

$\frac{P_{1}}{n_{1} \cdot T_{1}} = \frac{P_{2}}{n_{2} \cdot T_{2}}$ $P_1/(n_1*T_1) = P_2/(n_2*T_2)$

however, the number of moles is constant as well, given the fact that the gas is simply heated, so the ecuation becomes

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ $P_1/T_1 = P_2/T_2$

This will result in $T_{2} = \frac{P_{2}}{P_{1}} \cdot T_{1}$ $T_2=P_2/P_1 * T_1$ = $\frac{0.880}{0.325} \cdot 298 = 806.9 K$ $0.880/0.325 * 298=806.9 K$

keep in mind the temperature must be transformed in Kelvins by adding 273.15 (I just added 273 for simplicity)...

Finally, check to see if the answer make sense...the pressure is increasing in a constant volume, so the temperature MUST be bigger than $T_{1}$ $T_1$ ...

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Question #c6627

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