In one vessel, the volume of which V1 = 1.6 l, there is m1

One vessel with a volume V1 = 1.6 l contains m1 = 14 mg of nitrogen, another vessel with a volume V2 = 3.40 l contains m2 = 16 mg of oxygen at equal temperatures. After the vessels are connected and the gases are mixed, it is necessary to find the increase in entropy during this process.

To find the entropy increment, it is necessary to use the formula ΔS = Cv * ln(T2/T1) + R * ln(V2/V1), where Cv is the heat capacity of the gas at constant volume, R is the universal gas constant, T1 and T2 are the initial and final temperatures gases, V1 and V2 - initial and final volumes of gases.

Since the temperatures of the gases are equal, the first term in the formula becomes zero. Substituting the values ​​of volumes and masses of gases, we obtain ΔS = 5.7 J/K.

Thus, when gases are mixed, the entropy increases by 5.7 J/K.

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In one vessel, the volume of which V1 = 1.6 l, there is m1 mass of this digital product.

Product description: In one vessel, the volume of which V1 = 1.6 l, there is m1 mass of this digital product.

Physics problem: One vessel with a volume V1=1.6 l contains m1=14 mg of nitrogen, another vessel with a volume V2=3.40 l contains m2=16 mg of oxygen at equal temperatures. After the vessels are connected and the gases are mixed, it is necessary to find the increase in entropy during this process.

To solve the problem, use the formula ΔS = Cv * ln(T2/T1) + R * ln(V2/V1), where Cv is the heat capacity of the gas at constant volume, R is the universal gas constant, T1 and T2 are the initial and final temperatures of the gases, V1 and V2 - initial and final volumes of gases.

Since the temperatures of the gases are equal, the first term in the formula becomes zero. Substituting the values ​​of volumes and masses of gases, we obtain ΔS = 5.7 J/K.

Thus, when gases are mixed, the entropy increases by 5.7 J/K. Answer: ΔS = 5.7 J/K.

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To solve problem 20064 it is necessary to use the entropy increment formula for an ideal gas:

ΔS = C_p ln(T2/T1) - R ln(V2/V1)

where ΔS is the entropy increment, C_p is the heat capacity at constant pressure, R is the universal gas constant, T1 and T2 are the gas temperatures before and after mixing, V1 and V2 are the volumes of vessels before and after mixing.

In this problem, the gas temperatures are equal, so the first term in the entropy increment formula is zero. The volumes of the vessels and the masses of gases are also known, so their densities can be expressed:

ρ1 = m1/V1 ρ2 = m2/V2

After mixing, the gases are evenly distributed throughout both vessels, so the final density of the gases can be expressed:

ρ = (m1 + m2) / (V1 + V2)

Thus, we can calculate the entropy increment:

ΔS = R ln(ρ/ρ1) + R ln(ρ/ρ2)

Substituting numerical values, we get:

ΔS ≈ 2.8 J/K

Answer: the increase in entropy when mixing gases is approximately 2.8 J/K.

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