Question

1. a helium balloon with a pressure of 1.00 atm and a volume of 4.50 l at 20.0 °c is released into the atmosphere. what volume will the balloon occupy at an altitude where the pressure is 0.600 atm and the temperature is -20.0 °c? p = 1.00 p2 = 0.6 v = 4.5 v2 =? t = 293k t2 = 253k $\frac{1.00(4.5)}{293}=\frac{0.6(x)}{253}$ 4.5 0.6x 37.5 75.2

Explanation:

Step1: Convert temperatures to Kelvin

$T_1 = 20.0 + 273 = 293\ K$
$T_2=-20.0 + 273 = 253\ K$

Step2: Apply the combined - gas law $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

We know $P_1 = 1.00\ atm$, $V_1 = 4.50\ L$, $T_1 = 293\ K$, $P_2 = 0.600\ atm$, $T_2 = 253\ K$. Rearranging for $V_2$ gives $V_2=\frac{P_1V_1T_2}{P_2T_1}$.

Step3: Substitute the values into the formula

$V_2=\frac{1.00\ atm\times4.50\ L\times253\ K}{0.600\ atm\times293\ K}$
$V_2=\frac{4.50\times253}{0.600\times293}$
$V_2=\frac{1138.5}{175.8}$
$V_2 = 6.48\ L$

Answer:

$6.48\ L$