If a balloon containing 1.200 L of gas at 25°C and 760 mmHg pressure ascends to an altitude where the pressure is 380 and the temperature is 54°C, what will the volume be?

Jan 12, 2017

We use the old relationship: $\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

Explanation:

Of course, we must use values of $\text{absolute temperature}$. Given the proportionality we could units of pressure and volume of $\text{foot pounds}$ or $\text{atmospheres}$ or $\text{gallons}$ so long as we are consistent. Of course, I will use units of $\text{litres}$, and $\text{atmospheres}$.

${V}_{2} = \frac{{P}_{1} {V}_{1}}{T} _ 1 \times {T}_{2} / {P}_{2} =$

$\frac{1.200 \cdot L \times 1 \cdot a t m}{293 \cdot K} \times \frac{327 \cdot K}{\frac{380 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1}} \cong 2.7 \cdot L$

But don't trust my arithmetic. You have to do this on paper, and make sure all the given units cancel out to give an answer in $\text{litres}$ as required for a volume. Note that since we decrease the pressure, and INCREASE the temperature, the rise in volume is consistent with our expectation.