Hikers noticed that a sealed bag of potato chips puffs up when taken to the top of a mountain. Suppose that at the valley floor below, the air pressure is 1.0 atm, the temperature is 25^circ C and the volume of the bag is 0.985 L. At the top of the mountain, the temperature is 22^circ C and the bag has puffed up to 1.030 L What is the air pressure on top of the mountain? Type in your answer using the correct number of significant figures. Formula: (P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)) square atm

To solve this problem, we can use the ideal gas law equation:<br /><br />$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$<br /><br />where:<br />- $P_{1}$ is the initial pressure (1.0 atm)<br />- $V_{1}$ is the initial volume (0.985 L)<br />- $T_{1}$ is the initial temperature (25°C + 273.15 = 298.15 K)<br />- $P_{2}$ is the final pressure (unknown)<br />- $V_{2}$ is the final volume (1.030 L)<br />- $T_{2}$ is the final temperature (22°C + 273.15 = 295.15 K)<br /><br />Plugging in the values, we get:<br /><br />$\frac{1.0 \text{ atm} \times 0.985 \text{ L}}{298.15 \text{ K}} = \frac{P_{2} \times 1.030 \text{ L}}{295.15 \text{ K}}$<br /><br />Solving for $P_{2}$, we get:<br /><br />$P_{2} = \frac{1.0 \text{ atm} \times 0.985 \text{ L} \times 295.15 \text{ K}}{298.15 \text{ K} \times 1.030 \text{ L}}$<br /><br />$P_{2} \approx 0.999 \text{ atm}$<br /><br />Therefore, the air pressure on top of the mountain is approximately 1.0 atm.