What volume of Ar at 45°C and 1.25 atm contains the same number of particles as 0.655 L of H₂ at 15°C and 2.00 atm?

Question

Quizplus · Accepted Answer

To find the volume of Ar that contains the same number of particles as the given volume of another gas, we can use the ideal gas law equation in the form of $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is temperature in Kelvin. Since the number of particles (moles) is the same for both gases, we can set up a ratio using the conditions for each gas and solve for the unknown volume of Ar.First, convert temperatures to Kelvin:- For H₂: $T_1 = 15°C + 273.15 = 288.15 K$- For Ar: $T_2 = 45°C + 273.15 = 318.15 K$Using the equation $P_1V_1/T_1 = P_2V_2/T_2$, where $P_1 = 2.00 atm$, $V_1 = 0.655 L$, $T_1 = 288.15 K$, $P_2 = 1.25 atm$, and $T_2 = 318.15 K$, we solve for $V_2$, the volume of Ar.Rearranging the equation gives $V_2 = (P_1V_1T_2)/(P_2T_1)$.Plugging in the values gives $V_2 = (2.00 atm * 0.655 L * 318.15 K) / (1.25 atm * 288.15 K)$, which calculates to approximately 1.16 L.

What Volume of Ar at 45°C and 1