Answer
882.3 K
Work Step by Step
Here we use the ideal gas law $PV=nRT$ to find the temperature of the neon.
$PV=nRT=\{\frac{mass(m)}{molecular\space mass(M)}\}RT$
$P=(\frac{m}{V})(\frac{RT}{M})=\frac{\rho RT}{M}-(1)$
Given that,
$P_{He}=P_{Ne}$, Therefore (1)=>
$\frac{\rho RT_{He}}{M_{He}}=\frac{\rho RT_{Ne}}{M_{Ne}}=>T_{Ne}=\frac{M_{Ne}T_{He}}{M_{He}}$
Let's plug known values into this equation.
$$T_{Ne}=\frac{(20.18\space g/mol)(175\space K)}{4.0026\space g/mol}=882.3\space K$$
Temperature of Neon = 882.3 K
