# Question #7a48b

Apr 22, 2015

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

$P \cdot {V}_{1} = {n}_{1} \cdot R \cdot {T}_{1}$ $\textcolor{b l u e}{\left(1\right)}$

The final state of the balloon will have

$P \cdot {V}_{2} = {n}_{2} \cdot R \cdot {T}_{2}$ $\textcolor{b l u e}{\left(2\right)}$

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

${n}_{2} = {n}_{1} / 2$ and ${T}_{2} = 2 \cdot {T}_{1}$

Plug this into equation $\textcolor{b l u e}{\left(2\right)}$

$P \cdot {V}_{2} = {n}_{1} / 2 \cdot R \cdot 2 {T}_{1}$

Now divide equation $\textcolor{b l u e}{\left(1\right)}$ by equation $\textcolor{b l u e}{\left(2\right)}$ to get a relationship between the two volumes

$\frac{\textcolor{b l u e}{\left(1\right)}}{\textcolor{b l u e}{\left(2\right)}} \implies \left(\cancel{\text{P") * V_1)/(cancel("P") * V_2) = (cancel(n_1) * cancel("R") * cancel(T_1))/(cancel(n_1)/2 * cancel("R}} \cdot 2 \cancel{{T}_{1}}\right)$

${V}_{1} / {V}_{2} = \frac{1}{\frac{1}{2}} \cdot \frac{1}{2} = 2 \cdot \frac{1}{2} = 1$

This implies that

$\textcolor{g r e e n}{{V}_{2} = {V}_{1}}$

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