A cylinder with a volume of V=22.4 l contains hydrogen at normal

A cylinder with a volume of V=22.4 liters contains hydrogen under normal conditions. After adding a certain amount of helium to the balloon, the pressure in it increased to 0.25 MPa, but the temperature remained the same. It is necessary to determine the mass of helium introduced into the balloon.

To solve this problem, we use the Boyle-Marriott law, which states that at a constant temperature for a given amount of gas, the product of pressure and volume remains constant. We also use the equation of state of an ideal gas:

pV = nRT,

where p is the gas pressure, V is its volume, n is the amount of substance, R is the universal gas constant, T is temperature.

Let us rewrite the equation of state of the gas in the form:

n = pV / RT.

Since the temperature has not changed, the amount of substance in the cylinder remains the same. After adding helium to the balloon, the amount of substance became equal to the sum of the amounts of hydrogen and helium:

n = n(H2) + n(He).

Thus, based on the Boyle-Marriott law, we can create the equation:

p(H2)V = (n(H2) + n(He))RT.

Let's express the amount of helium:

n(He) = (p(H2)V - n(H2)RT) / RT.

Now we can calculate the mass of helium introduced into the balloon using the molar mass of helium:

m(He) = M(He) * n(He),

where M(He) is the molar mass of helium.

Answer:

m(He) = M(He) * (p(H2)V - n(H2)RT) / RT.

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Hopefully:

• cylinder volume V=22.4 l;
• pressure in the cylinder after adding helium p=0.25 MPa;
• the temperature remained the same;
• the cylinder initially contained hydrogen under normal conditions.

You need to find: the mass of helium introduced into the balloon.

Solution: To solve the problem, we use the Boyle-Marriott law: at a constant temperature for a given amount of gas, the product of pressure and volume remains constant. We also use the equation of state of an ideal gas: pV = nRT, where p is the gas pressure, V is its volume, n is the amount of substance, R is the universal gas constant, T is temperature. Let's rewrite the equation of state of the gas in the form: n = pV / RT. Since the temperature has not changed, the amount of substance in the cylinder remains the same. After adding helium to the balloon, the amount of substance became equal to the sum of the amounts of hydrogen and helium: n = n(H2) + n(He). Thus, based on the Boyle-Marriott law, we can create the equation: p(H2)V = (n(H2) + n(He))RT. Let's express the amount of helium: n(He) = (p(H2)V - n(H2)RT) / RT. Now we can calculate the mass of helium introduced into the balloon using the molar mass of helium: m(He) = M(He) * n(He), where M(He) is the molar mass of helium.

Let's substitute the known values ​​and solve the equation:

• cylinder volume V=22.4 l;
• pressure in the cylinder after adding helium p=0.25 MPa;
• the temperature remained the same, which means T=293 K (normal conditions);
• for hydrogen, the molar mass M(H2) = 2 g/mol;
• for helium, molar mass M(He) = 4 g/mol;
• universal gas constant R = 8.31 J/(mol*K).

First, let's find the amount of hydrogen substance in the cylinder: n(H2) = p(H2) * V / (R * T) = 101325 Pa * 0.0224 m³ / (8.31 J/(mol*K) * 293 K) = 0.902 mol.

Now let's find the amount of helium substance: n(He) = (p(H2) * V - n(H2) * R * T) / (R * T) = (0.25 MPa * 0.0224 m³ - 0.902 mol * 8.31 J/(molK) * 293 K) / (8.31 J/(molK) * 293 K) = 0.025 mol.

Finally, let's find the mass of helium: m(He) = M(He) * n(He) = 4 g/mol * 0.025 mol = 0.1 g.

Answer: the mass of helium introduced into the balloon is 0.1 g.

***

A cylinder with a volume of V = 22.4 liters contained hydrogen under normal conditions, that is, at a pressure of 101.325 Pa and a temperature of 273.15 K. After a certain amount of helium was additionally introduced into the cylinder, the pressure in the cylinder increased to 0.25 MPa , which is equal to 25000 Pa, and the temperature remained unchanged (273.15 K).

To solve the problem, we will use the Boyle-Marriott law, which states that at a constant temperature, the amount of gas contained in a vessel is inversely proportional to its pressure. We will also need the equation of state of an ideal gas, which allows us to relate the pressure, volume, temperature and amount of gas substance.

So, let m helium be introduced into the balloon. Then the total amount of gas substance in the cylinder will be equal to the amount of hydrogen, which is equal to the mass of hydrogen (let it be M) divided by its molar mass, plus the amount of helium substance, which is equal to the mass of helium (m) divided by its molar mass:

n = M/(2 g/mol) + m/(4 g/mol)

Here we took into account that the molar mass of hydrogen is 2 g/mol, and that of helium is 4 g/mol.

According to the ideal gas equation of state, pressure P, volume V and amount of substance n are related as follows:

PV = nR*T

Here R is the universal gas constant equal to 8.31 J/(mol*K), and T is the absolute temperature of the gas.

Using this formula for the initial conditions, we get:

101325 Pa * 22.4 l = M/(2 g/mol) * 8.31 J/(mol*K) * 273.15 K

From here we find M = 2 g/mol * 101325 Pa * 22.4 l / (8.31 J/(mol*K) * 273.15 K) = 2.02 kg

Now we can write the formula for pressure after adding helium:

P' = (M/(2 g/mol) + m/(4 g/mol)) * R * T / V

Substituting known values ​​into it and solving for m, we get:

m = 4 g/mol * V * (P' - P)/(R * T) = 4 g/mol * 22.4 l * (25000 Pa - 101325 Pa) / (8.31 J/(mol*K ) * 273.15 K) = 0.19 kg

Thus, the mass of helium introduced into the balloon is 0.19 kg.

***

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