Question

critical thinking

1. applying concepts what happens to the volume of a balloon that is taken outside on a cold winter day? explain.
2. making inferences when scientists record a gas’s volume, they also record its temperature and pressure. why?
3. analyzing ideas what happens to the pressure of a gas if the volume of gas is tripled at a constant temperature?

Explanation:

Question 5

This involves gas laws (like Charles's Law). Charles's Law states that at constant pressure, the volume of a gas is directly proportional to its absolute temperature ($V \propto T$ for constant $P$). When the balloon is taken outside on a cold day, the temperature ($T$) of the gas inside decreases. Since volume and temperature are directly related (at constant pressure, assuming the balloon is flexible and pressure outside is constant), as temperature decreases, the volume of the gas (and thus the balloon) decreases. So the balloon will shrink (decrease in volume).

The volume of a gas is affected by both temperature and pressure (as per gas laws like Boyle's Law - $P \propto \frac{1}{V}$ at constant $T$, Charles's Law - $V \propto T$ at constant $P$, and the Combined Gas Law). So, to accurately report a gas's volume, the conditions of temperature and pressure (which influence volume) must also be recorded because volume is not an independent property of a gas; it depends on $T$ and $P$. For example, if temperature increases, volume increases (at constant $P$), and if pressure increases, volume decreases (at constant $T$). Thus, without noting $T$ and $P$, the volume value is not fully meaningful or reproducible.

Step1: Recall Boyle's Law

Boyle's Law states that for a fixed amount of gas at constant temperature, the pressure ($P$) and volume ($V$) of the gas are inversely proportional. Mathematically, this is expressed as $P_1V_1 = P_2V_2$, where $P_1, V_1$ are initial pressure and volume, and $P_2, V_2$ are final pressure and volume.

Step2: Define the change in volume

Let the initial volume be $V_1 = V$ and the final volume be $V_2 = 3V$ (since the volume is tripled), and temperature $T$ is constant.

Step3: Apply Boyle's Law

From $P_1V_1 = P_2V_2$, we can solve for $P_2$. Substituting the values: $P_1 \times V = P_2 \times 3V$.

Step4: Solve for $P_2$

Divide both sides by $3V$: $P_2 = \frac{P_1V}{3V} = \frac{P_1}{3}$. So the pressure becomes one - third of its initial value.

Answer:

The volume of the balloon decreases. This is explained by Charles's Law (at constant pressure, volume of a gas is directly proportional to its absolute temperature). When the balloon is moved to a cold environment, the temperature of the gas inside drops, so its volume decreases (the balloon shrinks).

Question 6