P_(1)=498 T_(1)=21^circ C T_(2)=-27^circ C P_(2)=??kPa (You answer should be to the hundreths or 2 decimal places: 0 .00)

To solve this problem, we can use the ideal gas law in the form of the combined gas law, which relates pressure and temperature when volume and amount of gas are constant:<br /><br />\[<br />\frac{P_1}{T_1} = \frac{P_2}{T_2}<br />\]<br /><br />First, we need to convert the temperatures from degrees Celsius to Kelvin. The conversion formula is:<br /><br />\[<br />T(K) = T(^{\circ}C) + 273.15<br />\]<br /><br />So, for \(T_1\):<br /><br />\[<br />T_1 = 21 + 273.15 = 294.15 \, \text{K}<br />\]<br /><br />And for \(T_2\):<br /><br />\[<br />T_2 = -27 + 273.15 = 246.15 \, \text{K}<br />\]<br /><br />Now, substitute the known values into the combined gas law equation:<br /><br />\[<br />\frac{498}{294.15} = \frac{P_2}{246.15}<br />\]<br /><br />Solving for \(P_2\), we get:<br /><br />\[<br />P_2 = \frac{498 \times 246.15}{294.15}<br />\]<br /><br />Calculate \(P_2\):<br /><br />\[<br />P_2 = \frac{122595.7}{294.15} \approx 416.73 \, \text{kPa}<br />\]<br /><br />Therefore, the pressure \(P_2\) is approximately \(416.73 \, \text{kPa}\).