Question #7a48b

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Question #7a48b

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Answer 1 · 2015-04-22T08:31:28

The size of your balloon will remain unchanged.

This time, only pressure is kept constant. You can use the ideal gas law equation to determine what's going to happen to the balloon.

So, for the initial state, you'd get

$P \cdot V_{1} = n_{1} \cdot R \cdot T_{1}$ $P * V_1 = n_1 * R * T_1$ $(1)$ $color(blue)((1))$

The final state of the balloon will have

$P \cdot V_{2} = n_{2} \cdot R \cdot T_{2}$ $P * V_2 = n_2 * R * T_2$ $(2)$ $color(blue)((2))$

Since you know that the number of moles is halved and the temperature is doubled, you can write

$n_{2} = \frac{n_{1}}{2}$ $n_2 = n_1/2$ and $T_{2} = 2 \cdot T_{1}$ $T_2 = 2 * T_1$

Plug this into equation $(2)$ $color(blue)((2))$

$P \cdot V_{2} = \frac{n_{1}}{2} \cdot R \cdot 2 T_{1}$ $P * V_2 = n_1/2 * R * 2T_1$

Now divide equation $(1)$ $color(blue)((1))$ by equation $(2)$ $color(blue)((2))$ to get a relationship between the two volumes

$(color(blue)((1)))/(color(blue)((2))) => (cancel("P") * V_1)/(cancel("P") * V_2) = (cancel(n_1) * cancel("R") * cancel(T_1))/(cancel(n_1)/2 * cancel("R") * 2cancel(T_1))$

$V_1/V_2 = 1/(1/2) * 1/2 = 2 * 1/2 = 1$

This implies that

$color(green)(V_2 = V_1)$

The decrease in volume caused by the fact that you have less moles of gas is compensated by the doubling in temperature.

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