A balloon is taken outside on a hot summer day. With an increase in temperature, one should expect the pressure in the balloon to Increase Decrease stay the same

To solve this problem, we can apply the ideal gas law, which states that: \[ PV = nRT \] Where: - \( P \) is the pressure, - \( V \) is the volume, - \( n \) is the number of moles of gas, - \( R \) is the ideal gas constant, and - \( T \) is the temperature in Kelvin. ### Step-by-step Analysis: 1. **Understanding the Variables**: - When the balloon is taken outside on a hot summer day, the temperature \( T \) increases. - Assuming the volume \( V \) of the balloon remains constant (as the balloon expands but is still constrained by its material), we can analyze how pressure \( P \) changes with temperature. 2. **Rearranging the Ideal Gas Law**: - If we rearrange the ideal gas law to focus on pressure, we get: \[ P = \frac{nRT}{V} \] - Here, if \( n \) (the amount of gas) and \( V \) (the volume of the balloon) are constant, then pressure \( P \) is directly proportional to temperature \( T \). 3. **Conclusion**: - As the temperature \( T \) increases, the pressure \( P \) must also increase, assuming the volume of the balloon does not change significantly. ### Final Answer: The pressure in the balloon will **increase** with an increase in temperature.