Two flasks I and II shown below are connected by a valve of negligible volume. When the valve is ope

Q: Why do we use $C_v$ instead of $C_p$ when calculating the heat transfer?

We use $C_v$ because the internal energy of an ideal gas is strictly a function of temperature, given by $U = n C_v T$, regardless of whether the process is at constant volume or constant pressure. The total internal energy of the isolated system is conserved.

Q: Would the final temperature be different if one flask had Helium and the other had Nitrogen?

Yes, absolutely! Helium is monoatomic ($C_v = \frac{3}{2}R$) and Nitrogen is diatomic ($C_v = \frac{5}{2}R$). The $C_v$ terms would no longer cancel out in the energy conservation equation, shifting the final equilibrium temperature.

Q: Why did we convert the volume from liters to cubic meters?

The universal gas constant $R$ is given as $8.31\text{ J mol}^{-1}\text{ K}^{-1}$. Joules are SI units, which require volume to be in cubic meters ($m^3$) and pressure to be in Pascals ($Pa$) to maintain dimensional consistency.

Q: Can we just use Boyle's Law ($P_1 V_1 + P_2 V_2 = P_{final} V_{final}$) here?

No, Boyle's Law only applies when the temperature is constant throughout the entire process. Here, the two gases start at drastically different temperatures ($300\text{ K}$ and $60\text{ K}$), so temperature is a dynamic variable that must be accounted for.

Q: Why is the total volume simply the sum of the two volumes? Doesn't the gas compress?

The volume of a gas is defined by the container it occupies. Once the valve is opened, the gas molecules are free to move throughout both flasks. The rigid walls of the flasks don't change size, so the total available space is exactly $V_1 + V_2$.

Question

Two flasks I and II shown below are connected by a valve of negligible volume.

When the valve is opened, the final pressure of the system in bar is $x 	imes 10^{-2}$. The value of $x$ is ............. . (Integer answer)
[Assume, Ideal gas, $1	ext{ bar} = 10^5	ext{ Pa}$, molar mass of $	ext{N}_2 = 28.0	ext{ g mol}^{-1}$; $R = 8.31	ext{ J mol}^{-1}	ext{ K}^{-1}$]

Sigma Percentile · Accepted Answer

The problem of mixing gases from two different flasks is a classic application of the First Law of Thermodynamics and the Ideal Gas Law. It tests your ability to track the conservation of energy and apply state equations to a combined system. Let's break down the physics behind this process.

### Analyzing the Setup

We are given two flasks, I and II, connected by a valve. Before the valve is opened, the gases are isolated from each other.

In Flask I, we have $2.8	ext{ g}$ of Nitrogen gas ($N_2$) at a temperature of $300	ext{ K}$ in a volume of $1	ext{ L}$. 
In Flask II, we have $0.2	ext{ g}$ of Nitrogen gas at $60	ext{ K}$ in a volume of $2	ext{ L}$.

First, let's convert the given masses into moles, as the ideal gas law operates on molar quantities. The molar mass of $N_2$ is $28	ext{ g mol}^{-1}$.

For Flask I:
$$n_A = \frac{2.8}{28} = 0.1	ext{ mol}$$

For Flask II:
$$n_B = \frac{0.2}{28} = \frac{1}{140}	ext{ mol}$$

### The Master Equation: Conservation of Energy

When the valve is opened, the gases mix. Because Flask I is much hotter ($300	ext{ K}$) than Flask II ($60	ext{ K}$), heat will naturally flow from the hotter gas to the colder gas until they reach a common equilibrium temperature, $T$.

Crucially, the problem implies that the entire two-flask system is thermally insulated from the outside world. This means the total internal energy of the system must remain constant.

$$\Delta U_{system} = 0$$
$$\Delta U_A + \Delta U_B = 0$$

This translates to a simple principle: the heat lost by the gas in Flask I is exactly equal to the heat gained by the gas in Flask II.

For an ideal gas, the change in internal energy is given by $\Delta U = n C_v \Delta T$. Since both flasks contain Nitrogen, a diatomic gas, they share the same molar heat capacity at constant volume, $C_v = \frac{5}{2} R$.

Equating the heat transfer:
$$n_A C_v (T_1 - T) = n_B C_v (T - T_2)$$

Notice how the $C_v$ terms beautifully cancel out from both sides! This leaves us with a straightforward algebraic equation:
$$n_A (T_1 - T) = n_B (T - T_2)$$

### Finding the Equilibrium Temperature

Let's substitute our known values into this relation:
$$0.1 (300 - T) = \frac{1}{140} (T - 60)$$

To make the algebra cleaner, let's multiply the entire equation by $140$:
$$14 (300 - T) = T - 60$$
$$4200 - 14T = T - 60$$

Bringing the temperature terms to one side:
$$15T = 4260$$
$$T = 284	ext{ K}$$

The final equilibrium temperature of the mixed gas is $284	ext{ K}$.

### Final Calculation: The Mixed Gas Pressure

Now that the gases are fully mixed and at thermal equilibrium, we can treat them as a single, unified system.

The total volume is simply the sum of the individual flask volumes:
$$V_{total} = 1	ext{ L} + 2	ext{ L} = 3	ext{ L} = 3 	imes 10^{-3}	ext{ m}^3$$

The total number of moles is the sum of the moles from each flask:
$$n_{total} = 0.1 + \frac{1}{140} = \frac{14}{140} + \frac{1}{140} = \frac{15}{140} = \frac{3}{28}	ext{ mol}$$

We can now apply the Ideal Gas Law, $PV = nRT$, to find the final pressure:
$$P_{final} = \frac{n_{total} R T_{final}}{V_{total}}$$

Substituting our values:
$$P_{final} = \frac{\left(\frac{3}{28}ight) 	imes 8.31 	imes 284}{3 	imes 10^{-3}}$$

Notice how the $3$ in the numerator and denominator cancel out perfectly:
$$P_{final} = \frac{8.31 	imes 284}{28} 	imes 10^3	ext{ Pa}$$
$$P_{final} = 84.28 	imes 10^3	ext{ Pa}$$

The question asks for the pressure in bar. Since $1	ext{ bar} = 10^5	ext{ Pa}$, we multiply our result by $10^{-5}$:
$$P_{final} = 84.28 	imes 10^3 	imes 10^{-5}	ext{ bar}$$
$$P_{final} = 84.28 	imes 10^{-2}	ext{ bar}$$

The problem states the final pressure is $x 	imes 10^{-2}	ext{ bar}$ and asks for the nearest integer value of $x$. Rounding $84.28$ to the nearest integer, we get **$84$**.

Enter Numerical Value:

Visualized Solution

$n_{A} and n_{B}$

$Δ U_{sy s t e m} = 0$

$n_{A} C_{v} Δ T_{A} = n_{B} C_{v} Δ T_{B}$

$T_{f ina l}$

$n_{t o t a l} and V_{t o t a l}$

$P_{f ina l}$

$x value$

$Different Gases?$

The Sigma Insight: Gaseous State

Analyzing the Setup

The Master Equation

Finding the Equilibrium Temperature

Final Calculation

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Enter Numerical Value:

Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔUsystem​=0

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tfinal​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ntotal​ and Vtotal​

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x value

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Different Gases?