Question

If the temperature of the helium balloon were increased from 30°C to 35°C and the volume of the balloon only expanded from 0.47L to 1.55L, how much pressure would be inside the balloon?

Answer 1

Answer:

Final pressure will be 0.308 times the initial pressure

Explanation:

To solve this problem, we can use the equation of state for an ideal gas:

$pV=nRT$

where

p is the gas pressure

V is the volume of the gas

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

For a gas under transformation, we can write the equation as:

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where for the gas in this problem:

$p_1$ = is the initial pressure (not given)

$V_1=0.47 L$ is the initial volume

$T_1=30C+273=303 K$ is the initial temperature

$p_2 =$ is the final pressure

$V_2=1.55 L$ is the final volume

$T_2=35+273 = 308 K$ is the final temperature

Solving for p2, we find:

$p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{p_1 (0.47)(308)}{(303)(1.55)}=0.308p_1$