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Chapter 5: Problem 139

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

Short Answer

Expert verified
1492 balloons can be inflated to a pressure of 1.00 atm using the gas from the tank.

Step by step solution

01

Write down the initial conditions of the gas in the tank

The initial conditions of the gas in the tank are as follows: - Volume (V1): 15.0 L - Pressure (P1): \(2.00 \times 10^2\) atm Additionally, we should note that the tank cannot be emptied below 1.00 atm pressure.
02

Write down the ideal gas law equation for the given conditions

In this step, we will use the ideal gas law equation which states that for any gas, \(P_1V_1 = P_2V_2\), where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume, respectively.
03

Calculate the final conditions of the gas in the tank

As mentioned, we cannot empty the tank below 1.00 atm pressure. Therefore, we have: - Final Pressure (P2): 1.00 atm Now, we can re-arrange the ideal gas law equation to find the final volume (V2): \(V_2 = \frac{P_1V_1}{P_2}\) Substitute the given values, and find V2: \(V_2 = \frac{(2.00 \times 10^2 \mathrm{atm})(15.0 \mathrm{L})}{1.00 \mathrm{atm}} = 3000 \mathrm{L}\)
04

Calculate the volume of gas available for inflating the balloons

Now, find the volume available for inflating the balloons by taking the difference between V2 and the initial volume of the tank (V1): \(V_\mathrm{available} = V_2 - V_1 = 3000 \mathrm{L} - 15.0 \mathrm{L} = 2985 \mathrm{L}\)
05

Divide the available volume of the gas by the volume of one balloon

Since each balloon needs to be inflated to a volume of 2.00 L at a pressure of 1.00 atm, we can find the total number of balloons: \(N_\mathrm{balloons} = \frac{V_\mathrm{available}}{V_\mathrm{balloon}}\) Substitute the values and find the number of balloons: \(N_\mathrm{balloons} = \frac{2985 \mathrm{L}}{2.00 \mathrm{L}} = 1492.5\) Since we cannot have a fraction of a balloon, we have to round down the number of balloons to the nearest whole number. Therefore, the final answer is:
06

Final answer

1492 balloons can be inflated to a pressure of 1.00 atm using the gas from the tank.

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Most popular questions from this chapter

In 1897 the Swedish explorer Andreé tried to reach the North Pole in a balloon. The balloon was filled with hydrogen gas. The hydrogen gas was prepared from iron splints and diluted sulfuric acid. The reaction is $$\mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{FeSO}_{4}(a q)+\mathrm{H}_{2}(g)$$ The volume of the balloon was 4800 \(\mathrm{m}^{3}\) and the loss of hydrogen gas during filling was estimated at \(20 . \%\) . What mass of iron splints and 98\(\%\) (by mass) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) were needed to ensure the complete filling of the balloon? Assume a temperature of $0^{\circ} \mathrm{C},\( a pressure of 1.0 atm during filling, and 100\)\%$ yield.

A container is filled with an ideal gas to a pressure of 11.0 atm at \(0^{\circ} \mathrm{C} .\) a. What will be the pressure in the container if it is heated to $45^{\circ} \mathrm{C} ?$ b. At what temperature would the pressure be 6.50 atm? c. At what temperature would the pressure be 25.0 atm?

A steel cylinder contains 5.00 mole of graphite (pure carbon) and 5.00 moles of \(\mathrm{O}_{2} .\) The mixture is ignited and all the graphite reacts. Combustion produces a mixture of \(\mathrm{CO}\) gas and \(\mathrm{CO}_{2}\) gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by 17.0\(\%\) . Calculate the mole fractions of \(\mathrm{CO}, \mathrm{CO}_{2},\) and \(\mathrm{O}_{2}\) in the final gaseous mixture.

You have a helium balloon at 1.00 atm and \(25^{\circ} \mathrm{C} .\) You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is 79.0\(\%\) nitrogen and 21.0\(\%\) oxygen by volume. The “lift” of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than \(25^{\circ} \mathrm{C} ?\) Explain. b. Calculate the temperature of the air required for the hot-air balloon to provide the same lift as the helium balloon at 1.00 atm and $25^{\circ} \mathrm{C}\( . Assume atmospheric conditions are 1.00 atm and \)25^{\circ} \mathrm{C} .$

A 2.50-L container is filled with 175 g argon. a. If the pressure is 10.0 atm, what is the temperature? b. If the temperature is 225 K, what is the pressure?
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