**Known:**
$V_1 = 10.0 text{L}$
$P_2 = 3P_1$
$T_2 = T_1 + 0.80T_1 = 1.80T_1$

**Unknown:**
$V_2 = ?$

**Calculation:**
To solve for the final volume, we use the Combined Gas Law

$$
frac{P_1V_1}{T_1} = frac{P_2V_2}{T_2}
$$

Isolating $V_2$ on one side of the equation

$$
V_2 = frac{P_1V_1T_2}{P_2T_1}
$$

Plugging in the given values, we have

$$
V_2 = frac{(cancel{P_1}) cdot (10) cdot (1.80cancel{T_1})}{(3cancel{P_1})cdot (cancel{T_1)}}
$$

$$
V_2 = 6.00 text{L}
$$

Thus, the answer is $boxed{textbf{B.} 6.00 text{L}}$

$textbf{B.} 6.00 text{L}$

In this problem, we want to find the temperature of nitrogen gas. We know at what pressure the gas is, and what is its volume. Also, we know the number of mols. To solve this task, we are going to use the ideal gas equation
$$
begin{aligned}
pV=nRT
end{aligned}
$$
where $p$ is a pressure, $V$ is a volume, $n$ is a number of mols, $R$ is the universal gas constant, and $T$ is a temperature.

First, we are going to find the relation for the temperature from the ideal gas equation
$$
begin{aligned}
pV&=nRT/:nR\
T&=frac{pV}{nR}
end{aligned}
$$

The pressure in this task is standard atmospheric pressure, so $p=101.3 hspace{0.5mm} mathrm{kPa}=101.3cdot 10^{3} hspace{0.5mm} mathrm{Pa}$. The gas has a volume of $V=0.080 hspace{0.5mm} mathrm{m^{3}}$, and the number of mols is $n=3.6 hspace{0.5mm} mathrm{mol}$. Also, we know the value of the universal gas constant $R=8.314 hspace{0.5mm} mathrm{frac{m^{3}Pa}{K cdot mol}}$.

Now, we can find the temperature
$$
begin{aligned}
T&=frac{pV}{nR}\
T&=frac{101.3cdot 10^{3} hspace{0.5mm} mathrm{Pa} cdot 0.080 hspace{0.5mm} mathrm{m^{3}}}{3.6 hspace{0.5mm} mathrm{mol} cdot 8.314 hspace{0.5mm} mathrm{frac{m^{3}Pa}{K cdot mol}}}\
T&=270.8 hspace{0.5mm} mathrm{K}\
T&approx 270hspace{0.5mm} mathrm{K}
end{aligned}
$$

$b)$ $270 hspace{0.5mm} mathrm{K}$

In this problem, we want to find the applied pressure. To do this, we can use the equation
$$
begin{aligned}
p=frac{F}{A}
end{aligned}
$$
where $p$ is a pressure, $F$ is an applied force, and $A$ is an area on which the force acts.

We know that the applied force is $F=200.0hspace{0.5mm} mathrm{N}$, and the area is $A=5.4hspace{0.5mm} mathrm{cm^{2}}=5.4cdot 10^{-4} hspace{0.5mm} mathrm{m^{2}}$.

Now, we can find the pressure
$$
begin{aligned}
p&=frac{F}{A}\
p&=frac{200.0hspace{0.5mm} mathrm{N}}{5.4cdot 10^{-4} hspace{0.5mm} mathrm{m^{2}}}\
p&=3.7 cdot 10^{5} hspace{0.5mm} mathrm{Pa}
end{aligned}
$$

$d)$ $3.7 cdot 10^{5} hspace{0.5mm} mathrm{Pa}$

In this problem, we need to find the area of the second piston. From the previous problem, we know the pressure is $p=3.7 cdot 10^{5} hspace{0.5mm} mathrm{Pa}$, and we know the equation
$$
begin{aligned}
p=frac{F}{A}
end{aligned}
$$
where $p$ is a pressure, $F$ is a force, and $A$ is an area.

First, we need to find the relation for an area
$$
begin{aligned}
p&=frac{F}{A}\
A&=frac{F}{p}
end{aligned}
$$

We know the exerted force is $F=41 000hspace{0.5mm} mathrm{N}$, and we can find the area
$$
begin{aligned}
A&=frac{F}{p}\
A&=frac{41 000hspace{0.5mm} mathrm{N}}{3.7 cdot 10^{5} hspace{0.5mm} mathrm{Pa}}\
A&=0.11hspace{0.5mm} mathrm{m^{2}}
end{aligned}
$$

$c)$ $0.11hspace{0.5mm} mathrm{m^{2}}$

In this problem, we have a body underwater. On that body, there are two forces that act on it. The gravity force $F_{g}$ and buoyant force $F_{b}$. We want to find the apparent weight, so we are going to use equation
$$
begin{aligned}
F_{app}&=F_{g}-F_{b}
end{aligned}
$$

The buoyant force can be written as the product of the volume of submerging part of a body, gravity acceleration, and a density of the fluid
$$
begin{aligned}
F_{b}&=rho_{w}hspace{0.5mm}g hspace{0.5mm}V_{sub}
end{aligned}
$$

The whole body is underwater, so the volume of the body underwater is the same as the whole body volume, $V_{sub}=V$, so we can write
$$
begin{aligned}
F_{b}&=rho_{w}hspace{0.5mm}g hspace{0.5mm}V
end{aligned}
$$

The gravity force can be written as
$$
begin{aligned}
F_{g}&=mg\
F_{g}&=Vrho_{wood}g\
end{aligned}
$$
where we know the density of cocobolo wood, $rho_{wood}=1.10 hspace{0.5mm}mathrm{g/cm^{3}}=1.10 cdot 10^{3} hspace{0.5mm}mathrm{kg/m^{3}}$. Also, we know the volume of the body,
$$
begin{aligned}
V&=786hspace{0.5mm}mathrm{mL}\
V&=786 cdot 10^{-3}hspace{0.5mm}mathrm{L}\
V&=786 cdot 10^{-3}hspace{0.5mm}mathrm{dm^{3}}\
V&=786 cdot 10^{-6}hspace{0.5mm}mathrm{m^{3}}
end{aligned}
$$

Now, we can find the apparent weight
$$
begin{aligned}
F_{app}&=F_{g}-F_{b}\
F_{app}&=Vrho_{wood}g-rho_{w}g V\
F_{app}&=Vg(rho_{wood}-rho_{w})\
F_{app}&=786 cdot 10^{-6}hspace{0.5mm}mathrm{m^{3}} cdot 9.8 hspace{0.5mm} mathrm{m/s^{2}}(1.10 cdot 10^{3} hspace{0.5mm}mathrm{kg/m^{3}}-10^{3} hspace{0.5mm}mathrm{kg/m^{3}})\
F_{app}&=0.770hspace{0.5mm}mathrm{N}
end{aligned}

$a)$ $0.770hspace{0.5mm}mathrm{N}$

In this problem, we need to find the buoyant force. The buoyant force can be written as the product of the volume of submerging part of a body, gravity acceleration, and a density of the fluid
$$
begin{aligned}
F_{b}&=rho_{fluid}hspace{0.5mm}g hspace{0.5mm}V_{sub}
end{aligned}
$$

In our case, the fluid is water, so the density is $rho=1000 hspace{0.5mm}mathrm{kg/m^{3}}$. The submerged volume is the same as the volume of displaced water, so
$$
begin{aligned}
V_{sub}&=85hspace{0.5mm}mathrm{L}\
V_{sub}&=85hspace{0.5mm}mathrm{dm^{3}}\
V_{sub}&=85cdot 10^{-3}hspace{0.5mm}mathrm{m^{3}}
end{aligned}
$$

Now, we can find the buoyant force
$$
begin{aligned}
F_{b}&=rhohspace{0.5mm}g hspace{0.5mm}V_{sub}\
F_{b}&=1000 hspace{0.5mm}mathrm{kg/m^{3}} cdot 9.8 hspace{0.5mm} mathrm{m/s^{2}} cdot 85cdot 10^{-3}hspace{0.5mm}mathrm{m^{3}}\
F_{b}&=833 hspace{0.5mm}mathrm{N}\
F_{b}&=8.3 cdot 10^{2} hspace{0.5mm}mathrm{N}
end{aligned}
$$

$b)$ $8.3 cdot 10^{2} hspace{0.5mm}mathrm{N}$

In this problem, we want to find the mass of carbon dioxide gas. We need to use two equations. First is the ideal gas equation
$$
begin{aligned}
pV=nRT
end{aligned}
$$
where $p$ is a pressure, $V$ is a volume, $n$ is a number of moles, $R$ is the universal gas constant, and $T$ is a temperature.
Also, we use the equation
$$
begin{aligned}
n=frac{m_{CO_{2}}}{M_{CO_{2}}}
end{aligned}
$$

We know the pressure is $p=3.0hspace{0.5mm} mathrm{atm}$. We can write this as $p=3.0 cdot 101.3 hspace{0.5mm} mathrm{kPa}=303.9cdot 10^{3} hspace{0.5mm} mathrm{Pa}$$. Also, we know the volume of the gas is
$$
begin{aligned}
V&=365hspace{0.5mm}mathrm{mL}\
V&=365 cdot 10^{-3}hspace{0.5mm}mathrm{L}\
V&=365 cdot 10^{-3}hspace{0.5mm}mathrm{dm^{3}}\
V&=365 cdot 10^{-6}hspace{0.5mm}mathrm{m^{3}}
end{aligned}
$$
The temperature of the gas is $t=24 degree C$ or $T=297 hspace{0.5mm} mathrm{K}$, and the molar mass for carbon dioxide is $M_{CO_{2}}=44.0 hspace{0.5mm} mathrm{frac{g}{mol}}$

Now, we can find the mass
$$
begin{aligned}
n&=frac{m_{CO_{2}}}{M_{CO_{2}}}\
m_{CO_{2}}&=nM_{CO_{2}}\
m_{CO_{2}}&=0.046hspace{0.5mm} mathrm{mol} cdot 44.0 hspace{0.5mm} mathrm{frac{g}{mol}}\
m_{CO_{2}}&=2.0hspace{0.5mm} mathrm{g}
end{aligned}
$$

$b)$ $2.0hspace{0.5mm} mathrm{g}$

In this problem, we need to find the new volume of the balloon at the depth of $h=1.27 hspace{0.5mm} mathrm{m}$. If this is the isothermal process, we are going to use Boyle’s Law, $pV=const.$. Also, we need to find the hydrostatic pressure at that depth, and we use the equation
$$
begin{aligned}
p=p_{0}+rho g h
end{aligned}
$$
where $p_{0}$ is an atmospheric pressure, $p_{0}=101.3 hspace{0.5mm} mathrm{kPa}=101.3 cdot 10^{3} hspace{0.5mm} mathrm{Pa}$, $rho$ is a density of the fluid, and $h$ is a depth of a point where we measure the pressure.

First, we are going to find the hydrostatic pressure
$$
begin{aligned}
p&=p_{0}+rho g h\
p&=101.3 cdot 10^{3} hspace{0.5mm} mathrm{Pa}+10^{3} hspace{0.5mm} mathrm{kg/m^{3}}cdot 9.8 hspace{0.5mm}mathrm{m/s^{2}} cdot 1.27 hspace{0.5mm} mathrm{m}\
p&=113.7 cdot 10^{3} hspace{0.5mm} mathrm{Pa}
end{aligned}
$$

Now, we can find the volume as
$$
begin{aligned}
p_{0}V&=pV’\
V’&=V frac{p_{0}}{p}\
V’&=125 hspace{0.5mm} mathrm{mL} frac{101.3 cdot 10^{3} hspace{0.5mm} mathrm{Pa}}{113.7 cdot 10^{3} hspace{0.5mm} mathrm{Pa}}
V’&=111.4hspace{0.5mm} mathrm{mL}
end{aligned}
$$

$V’=111.4hspace{0.5mm} mathrm{mL}$