At what final temperature, in degrees Celsius, will the pressure of the \( \mathrm{H}_{2} \) decrease to 0.27 atm, if volume and amount of gas do nof change? Express your answer using three significant figures.

To find the final temperature at which the pressure of \( \mathrm{H}_{2} \) decreases to 0.27 atm, we can use the ideal gas law in the form of \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), where \( P \) is pressure and \( T \) is temperature in Kelvin. Assuming the initial pressure \( P_1 \) is known (for this example, let's say it was 1 atm at \( T_1 = 273.15 \) K, which is 0°C): 1. We know that \( P_2 = 0.27 \) atm. 2. We want to find \( T_2 \). Rearranging the formula gives us: \( T_2 = \frac{P_2 \cdot T_1}{P_1} \). Substituting the values: \( T_2 = \frac{0.27 \, \text{atm} \cdot 273.15 \, \text{K}}{1 \, \text{atm}} \) \( T_2 ≈ 73.6935 \, \text{K} \). Now converting this back to degrees Celsius: \( T_2 (°C) = 73.6935 \, \text{K} - 273.15 ≈ -199.46 \, °C \). Finally, expressing this using three significant figures we have: **Final Temperature = -199 °C**.