A 15.0-L tank is filled with \(\mathrm{H}_{2}\) to a pressure of \(2.00 \times 10^{2}\) atm. How many balloons (each \(2.00 \mathrm{~L}\) ) can be inflated to a pressure of \(1.00\) atm from the tank? Assume that there is no temperature change and that the tank cannot be emptied below \(1.00\) atm pressure.

We are given the following information: Initial volume of the tank: \(V_1 = 15.0 L\) Initial pressure of the tank: \(P_1 = 2.00 \times 10^2 atm\) Final pressure of the tank: \(P_2 = 1.00 atm\) Final volume of each balloon: \(V_{balloon} = 2.00 L\) Additionally, we know that the temperature remains constant during the process. The Ideal Gas Law is given by the formula: \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles of the gas, \(R\) is the gas constant, and \(T\) is temperature. Since the temperature remains constant, we can relate the initial and final states using the following equation: \(\frac{P_1V_1}{n_1} = \frac{P_2V_2}{n_2}\).

Since we can't empty the tank below 1.00 atm pressure, we know that the final pressure in the tank will be 1.00 atm. Therefore, using the relationship \(\frac{P_1V_1}{n_1} = \frac{P_2V_2}{n_2}\), we can solve for the final number of moles in the tank, \(n_2\). \(n_2 = \frac{P_2V_2}{\frac{P_1V_1}{n_1}}\) Notice that \(V_2\) is the final volume of the tank, which is equal to the initial volume, \(V_1 = 15.0 L\). Thus, we can rewrite the equation as follows: \(n_2 = \frac{P_2V_1}{\frac{P_1V_1}{n_1}}\) Since the problem is only asking for the number of balloons that can be inflated, we don't actually need to know the number of moles. Rather, we can find the ratio of final moles to the initial moles.

Using the equation from the previous step, let's find the ratio \(r = \frac{n_2}{n_1}\): \(r = \frac{n_2}{n_1} = \frac{\frac{P_2V_1}{\frac{P_1V_1}{n_1}}}{n_1} = \frac{P_2V_1}{P_1V_1} = \frac{P_2}{P_1}\) Now, we can plug in the given pressures: \(r = \frac{1.00~atm}{2.00 \times 10^2~atm} = \frac{1}{200}\)

Since we now have the ratio of final moles to initial moles, we can find the number of moles that were used to inflate the balloons, \(n_{used} = n_1 - n_2\). Using the ratio from the previous step: \(n_{used} = n_1 - n_2 = n_1 - rn_1 = (1 - r)n_1 = (1-\frac{1}{200})n_1 = (\frac{199}{200})n_1\)

We are given that each balloon has a volume of 2.00 L and is inflated to a pressure of 1.00 atm. By applying the Ideal Gas Law again, we can find the number of moles required to inflate each balloon, \(n_{balloon}\): \(\frac{P_{balloon}V_{balloon}}{n_{balloon}} = \frac{P_1V_1}{n_1}\) We can solve for \(n_{balloon}\): \(n_{balloon} = \frac{P_{balloon}V_{balloon}}{\frac{P_1V_1}{n_1}} = \frac{P_{balloon}V_{balloon}}{P_1V_1}n_1\) Now, plug in the pressure and volume of each balloon: \(n_{balloon} = \frac{1.00~atm \cdot 2.00~L}{2.00 \times 10^2~atm \cdot 15.0~L}n_1 = \frac{1}{1.50\times10^3}n_1\) Finally, to find the number of balloons that can be inflated, divide the total number of moles used to inflate balloons (found in Step 4) by the number of moles required to inflate each balloon: \(N = \frac{n_{used}}{n_{balloon}} = \frac{(\frac{199}{200})n_1}{\frac{1}{1.50\times10^3}n_1} = \frac{199}{200} \cdot \frac{1}{\frac{1}{1.50\times10^3}} = \frac{199}{200} \cdot 1.50\times10^3 \approx 1.49\times10^3\) So, approximately 1,490 balloons can be inflated from the 15.0-L tank.