A 6.0-L tank is filled with helium gas at a pressure of 2 MPa. How many balloons (each \(2.00 \mathrm{~L}\) ) can be inflated to a pressure of $101.3 \mathrm{kPa}$, assuming that the temperature remains constant and that the tank cannot be emptied below \(101.3 \mathrm{kPa}\) ?

We are given: - Initial volume of the gas in the tank (V1) = 6.0 L - Initial pressure of the gas in the tank (P1) = 2 MPa = 2000 kPa - Volume of each balloon (V2) = 2.00 L - Pressure of gas in each balloon (P2) = 101.3 kPa Since the temperature remains constant, we will use Boyle's Law, which states that the product of the initial pressure and volume of gas is equal to the final pressure and volume (P1 × V1 = P2 × V2).

To calculate the maximum number of balloons, we will first determine the number of balloons that can be inflated without depleting the tank's pressure to below 101.3 kPa. First, calculate the final volume of the gas remaining in the tank (Vf = V1 × (P1 / P2)): \(V_f = V_1 \times (\frac{P_1}{P_2})\) Plugging in the given values: \(V_f = 6.0\,\text{L} \times (\frac{2000\text{kPa}}{101.3\text{kPa}})\) \(V_f \approx 118.56\,\text{L}\) Since the tank cannot deplete below 101.3 kPa, we need to ensure that at least 6 L of gas remains in the tank. So, the available volume of gas that can be used to inflate the balloons is: \(V_\text{usable} = V_f - V_1 = 118.56\,\text{L} - 6.0\,\text{L} = 112.56\,\text{L}\) Now, divide the available volume of gas by the volume of each balloon to find the number of balloons that can be inflated: \(\text{Number of balloons} = \frac{V_\text{usable}}{V_2} = \frac{112.56\,\text{L}}{2.00\,\text{L}} \approx 56.28\)

Since we cannot inflate a fraction of a balloon, we round down to the nearest whole number: Number of balloons = 56 Hence, 56 balloons can be inflated to a pressure of 101.3 kPa using the helium gas in the 6.0-L tank, without depleting the tank's pressure below 101.3 kPa.