Let's work through the problem step by step:

Given Data:

• Volume of the tank, $V = 0.025 \, \text{m}^3$
• Initial pressure, $P_1 = 200 \, \text{kPa}$
• Initial temperature, $T_1 = 25^\circ C = 298.15 \, \text{K}$
• The additional mass of helium added: $m = 1 \, \text{kg}$
• Rupture pressure: $P_{\text{rupture}} = 500 \, \text{kPa}$
• Molecular weight of helium: $M = 4.003 \, \text{kg/kmol}$
• The helium gas is assumed to behave as an ideal gas.

We need to find the new pressure after the gas has returned to $25^\circ C$ and check if it exceeds the rupture pressure.

Step 1: Calculate the initial moles of helium in the tank.

The number of moles of helium in the tank can be calculated using the ideal gas law: $P_1 V = n R T_1$ Where:

• $P_1 = 200 \, \text{kPa} = 200 \times 10^3 \, \text{Pa}$
• $V = 0.025 \, \text{m}^3$
• $T_1 = 298.15 \, \text{K}$
• $R = 8.314 \, \text{J/(mol·K)}$ (universal gas constant)

We can find the number of moles of helium before the additional helium is added using the rearranged ideal gas law: $n_1 = \frac{P_1 V}{R T_1}$

Substitute the values: $n_1 = \frac{(200 \times 10^3) \times 0.025}{8.314 \times 298.15}$ $n_1 = \frac{5000}{2477.9} \approx 2.02 \, \text{mol}$

Step 2: Find the mass of helium initially present.

We know the molecular weight of helium $M = 4.003 \, \text{kg/kmol}$, so the mass $m_1$ of helium in the tank initially is: $m_1 = n_1 \times M$ $m_1 = 2.02 \, \text{mol} \times \frac{4.003 \, \text{kg}}{\text{kmol}} \approx 0.00809 \, \text{kg}$

Step 3: Add the extra helium.

The additional helium added to the tank is $m_{\text{add}} = 1 \, \text{kg}$.

So, the total mass of helium after the addition is: $m_{\text{total}} = m_1 + m_{\text{add}} = 0.00809 \, \text{kg} + 1 \, \text{kg} = 1.00809 \, \text{kg}$

Step 4: Calculate the total moles of helium after the addition.

We can now calculate the total number of moles of helium after the extra mass is added: $n_{\text{total}} = \frac{m_{\text{total}}}{M}$ $n_{\text{total}} = \frac{1.00809}{4.003} \approx 0.252 \, \text{mol}$

Step 5: Calculate the new pressure using the ideal gas law.

The gas will return to the initial temperature of $25^\circ C$. Using the ideal gas law again, the new pressure $P_2$ can be calculated as: $P_2 = \frac{n_{\text{total}} R T_1}{V}$ Substituting the known values: $P_2 = \frac{0.252 \times 8.314 \times 298.15}{0.025}$ $P_2 \approx 448.1 \, \text{kPa}$

Step 6: Determine if the tank will rupture.

Since the calculated pressure $P_2 = 448.1 \, \text{kPa}$ is less than the rupture pressure of $500 \, \text{kPa}$, the tank will not rupture.

Conclusion:

• The new pressure after adding the extra helium is 448.1 kPa, which is below the rupture pressure of 500 kPa, so the tank will not rupture.

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Would you like a more detailed explanation of any part of the calculation? Let me know if you need further clarification!

Related Questions:

1. How does the ideal gas law apply to a non-ideal gas like helium under extreme conditions?
2. What are some methods to increase the volume of a gas tank to avoid high pressure?
3. How would the calculation change if the gas was not assumed to be ideal?
4. What is the relationship between molecular weight and the number of moles in a gas?
5. How does temperature influence the pressure of a gas when volume is held constant?

Tip: Always convert units appropriately before substituting values into formulas, especially when dealing with pressure, volume, and temperature!