Question

1. a toy balloon has an internal pressure of 1.05 atm and a volume of 5.0 l. if the temperature where the balloon is released is 20° c, what will happen to the volume when the balloon rises to an altitude where the pressure is 0.65 atm and the temperature is -15° c?
2. a small research submarine with a volume of 1.2 x 10^5 l has an internal pressure of 1.0 atm and an internal temperature of 15° c. if the submarine descends to a depth where the pressure is 150 atm and the temperature is 3° c, what will the volume of the gas inside be if the hull of the submarine breaks?
3. people who are angry sometimes say that they feel as if theyll explode. if a calm person with a lung capacity of 3.5 liters and a body temperature of 36° c gets angry, what will the volume of the persons lungs be if their temperature rises to 39° c. based on this, do you think its likely they will explode?

Explanation:

Step 1: Convert temperatures to Kelvin

For the initial state of the balloon in question 2: $T_1 = 20 + 273.15=293.15\ K$. For the final state: $T_2=- 15+273.15 = 258.15\ K$.
For the initial state of the submarine in question 3: $T_1 = 15+273.15=288.15\ K$. For the final state: $T_2=3 + 273.15=276.15\ K$.
For the initial state of the person in question 4: $T_1 = 36+273.15=309.15\ K$. For the final state: $T_2=39+273.15 = 312.15\ K$.

Step 2: Apply the combined - gas law $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$ and solve for $V_2$

Question 2:
Given $P_1 = 1.05\ atm$, $V_1 = 5.0\ L$, $P_2 = 0.65\ atm$, $T_1 = 293.15\ K$, $T_2 = 258.15\ K$.
Rearrange the combined - gas law to $V_2=\frac{P_1V_1T_2}{P_2T_1}$.
$V_2=\frac{1.05\times5.0\times258.15}{0.65\times293.15}$
$V_2=\frac{1.05\times5.0\times258.15}{0.65\times293.15}\approx6.9\ L$.
Question 3:
Given $P_1 = 1.0\ atm$, $V_1 = 1.2\times 10^{5}\ L$, $P_2 = 150\ atm$, $T_1 = 288.15\ K$, $T_2 = 276.15\ K$.
$V_2=\frac{P_1V_1T_2}{P_2T_1}$
$V_2=\frac{1.0\times1.2\times 10^{5}\times276.15}{150\times288.15}$
$V_2=\frac{1.0\times1.2\times 10^{5}\times276.15}{150\times288.15}\approx770\ L$.
Question 4:
Assume the pressure remains constant, so the gas law simplifies to $\frac{V_1}{T_1}=\frac{V_2}{T_2}$ (Charles's law). Given $V_1 = 3.5\ L$, $T_1 = 309.15\ K$, $T_2 = 312.15\ K$.
Rearrange to $V_2=\frac{V_1T_2}{T_1}$
$V_2=\frac{3.5\times312.15}{309.15}\approx3.53\ L$. Since the increase in volume is very small ($3.53 - 3.5=0.03\ L$), it is extremely unlikely that the lungs will explode.

Answer:

Question 2: The volume of the balloon will be approximately $6.9\ L$.
Question 3: The volume of the gas will be approximately $770\ L$.
Question 4: The volume of the lungs will be approximately $3.53\ L$. It is unlikely that they will explode.