Question #675cd

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Question #675cd

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Answer 1 · 2015-01-04T10:22:41

I assume the question refers to 4 L of gas, and asks for the new volume after the pressure and the temperature are doubled.

This can be solved by using the combined gas law,

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1 = (P_2V_2)/T_2$ .

You start with a pressure of $P_{1}$ $P_1$ and end with a pressure of $P_{2} = 2 \cdot P_{1}$ $P_2 = 2*P_1$ . Likewise, the initial temperature is $T_{1}$ $T_1$ , and the final temperature will be $T_{2} = 2 \cdot T_{1}$ $T_2 = 2 * T_1$ .

So, we can determine $V_{2}$ $V_2$ from the combined gas law equation

$V_{2} = \frac{P_{1}}{P_{2}} \cdot \frac{T_{2}}{T_{1}} \cdot V_{1} \Rightarrow V_{2} = \frac{P_{1}}{2 \cdot P_{1}} \cdot \frac{2 \cdot T_{1}}{T_{1}} \cdot V_{1}$ $V_2 = P_1/P_2 * T_2/T_1 * V_1 => V_2 = P_1/(2*P_1) * (2 * T_1)/T_1 * V_1$

It's evident that $V_{2} = V_{1}$ $V_2 = V_1$ , since both the pressure and the temperature terms cancel out.

This would have also been the case if the question said 4 moles of gas, since the combined gas law assumes that the number of moles is constant.

You would get the same result, $V_{final} = V_{initial}$ $V_("final") = V_("initial")$ .

SInce the question provides a value for $V_{1}$ $V_1$ , the answer is

$V_{2} = V_{1} = 4$ $V_2 = V_1 = 4$ $L$ $"L"$ .

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Question #675cd

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