We will need to use Boyle's Law to solve:

$\displaystyle P_1V_1=P_2V_2$

Boyle's Law allows us to set up a relationship between the changes in pressure and volume under conditions with constant temperature. Since the equation is a proportion, we do not need to convert any units.

We can use the given values to solve for the new pressure.

$\displaystyle (2Pa)(1L)=(P_2)(0.5L)$
 

$\displaystyle \frac{(2Pa)(1L)}{0.5L}=(P_2)$

$\displaystyle \frac{2Pa\cdot L}{0.5L}=P_2$

$\displaystyle 4Pa=P_2$

For this problem, use Charles's Law:

$\displaystyle \frac{V_1}{T_1}=\frac{V_2}{T_2}$

In this formula, $\displaystyle V$ is the volume and $\displaystyle T$ is the temperature. Charles's Law allows us to set up a proportion for changes in volume and temperature, as long as pressure remains constant. Since we are dealing with a proportion, the units for temperature are irrelevant and we do not need to convert to Kelvin.

Using the given values, we should be able to solve for the final temperature.

$\displaystyle \frac{V_1}{T_1}=\frac{V_2}{T_2}$

$\displaystyle \frac{5m^3}{60^oC}=\frac{3m^3}{T_2}$

Cross multiply.

$\displaystyle 5m^3* T_2=3m^3* 60^oC$

$\displaystyle 5m^3* T_2=180(^oC\cdot m^3)$

$\displaystyle T_2=\frac{180(^{\circ}C\cdot m^3)}{5m^3}$

$\displaystyle T_2=36^{\circ}C$

For this problem, use Boyle's Law: 

$\displaystyle P_1V_1=P_2V_2$

Boyle's Law allows us to set up a proportion between the pressure and volume at a constant temperature.

Using the values given, we can solve for the final pressure.

$\displaystyle P_1V_1=P_2V_2$

$\displaystyle 20Pa* 8L=P_2* 6L$

$\displaystyle 160Pa\cdot L=P_2* 6L$

$\displaystyle \frac{160Pa\cdot L}{6L}=P_2$

$\displaystyle 26.67Pa=P_2$

The volume of air in the balloon will increase when exposed to hotter temperatures, and decrease when exposed to colder temperatures. If we look at the ideal gas law, we can see that temperature and volume have a direct relationship. As one goes down, so does the other, assuming all other factors remain constant.

$\displaystyle PV=nRT$

We can also look at Charles's law of volumes:

$\displaystyle \frac{V_1}{T_1}=\frac{V_2}{T_2}$

The balloon is sealed, so the amount of gas in the balloon will not change, and the elasticity of the balloon means that pressure will also remain constant. As temperature decreases, volume must also decrease. Suppose that the temperature is halved in our question. The result would be half the volume, according to Charles's law.

$\displaystyle \frac{V_1}{T_1}=\frac{xV_2}{\frac{1}{2}T_2}=\frac{\frac{1}{2}V_2}{\frac{1}{2}T_2}$

By this logic, we can conclude that the balloon will shrink when placed in the cold water.

Heat is a form of energy. Adding heat to a gaseous system will increase the energy of the molecules, causing them to move faster and collide more frequently. This increased velocity results in the expansion of the gas.