Final Answer:
The new pressure of the radon gas, given a new volume of 1.78 L and a new temperature of 325 K, can be calculated using the combined gas law formula. The new pressure is approximately 1.44 atm.
Step-by-step explanation:
The combined gas law formula is expressed as \(P_1 \times V_1 / T_1 = P_2 \times V_2 / T_2\), where \(P_1\), \(V_1\), and \(T_1\) represent the initial pressure, volume, and temperature, respectively, and \(P_2\), \(V_2\), and \(T_2\) represent the final pressure, volume, and temperature, respectively.
Given initial conditions:
\(V_1 = 1.53 L\), \(P_1 = 1.15 atm\), \(T_1 = 305 K\)
And final conditions:
\(V_2 = 1.78 L\), \(T_2 = 325 K\)
Using the combined gas law formula:
\(P_2 = P_1 \times V_1 \times T_2 / (V_2 \times T_1)\)
Plugging in the values:
\(P_2 = 1.15 atm \times 1.53 L \times 325 K / (1.78 L \times 305 K)\)
\(P_2 \approx 1.44 atm\)
Therefore, the new pressure of the radon gas, considering the new volume and temperature, is approximately 1.44 atm. This calculation showcases how the combined gas law relates the initial and final conditions of pressure, volume, and temperature for a given gas sample. As the volume increases and temperature rises, the pressure of the gas also increases, as shown by the resulting pressure value obtained through the calculation.