Question

1. balloon in cold weather

a helium balloon has a volume of 2.0 l at 295k. on a winter day, the temperature drops to 283k. the balloon shrinks because the gas particles inside lose kinetic energy, move slower, and collide with the balloon walls with less force.
this phenomenon demonstrates the relationship between temperature and volume at constant pressure. outdoor balloon displays must take temperature changes into account, because cold weather can cause balloons to shrink or lose lift unexpectedly.
a. calculate the new volume at 283k

Explanation:

Step1: Apply Charles' Law

Charles' Law formula is $\frac{V_1}{T_1}=\frac{V_2}{T_2}$. Here, $V_1 = 2.0\ L$, $T_1=295\ K$, $T_2 = 283\ K$.

Step2: Solve for $V_2$

Rearrange the formula to $V_2=\frac{V_1\times T_2}{T_1}$. Substitute the values: $V_2=\frac{2.0\times283}{295}$.
Calculate $2.0\times283 = 566$, then $V_2=\frac{566}{295}\approx1.92\ L$.

Answer:

$1.92\ L$