Question

a gas has a density of 2.08 g/l at 48°c and p = 1.07 atm.
what will be its density at 157°c?
density at 157°c=
calculate the molar mass of the gas, assuming it obeys the ideal gas law.
molar mass=
g/mol
use the references to access important values if needed for this qu

Explanation:

Step1: Convert temperatures to Kelvin

For \(T_1 = 48^{\circ}C\), \(T_1=(48 + 273)K=321K\). For \(T_2 = 157^{\circ}C\), \(T_2=(157+ 273)K = 430K\).

Step2: Use the ideal - gas law and density formula

The ideal - gas law is \(PV=nRT=\frac{m}{M}RT\), and density \(
ho=\frac{m}{V}\). Rearranging \(PV=\frac{m}{M}RT\) gives \(
ho=\frac{PM}{RT}\). Since \(P\) and \(M\) are constant (\(
ho_1=\frac{PM}{RT_1}\), \(
ho_2=\frac{PM}{RT_2}\)), then \(\frac{
ho_2}{
ho_1}=\frac{T_1}{T_2}\).

Step3: Calculate the density at \(157^{\circ}C\)

Given \(
ho_1 = 2.08g/L\), \(T_1 = 321K\), \(T_2 = 430K\). Substitute into \(
ho_2=
ho_1\times\frac{T_1}{T_2}\). So \(
ho_2=2.08\times\frac{321}{430}\) \(g/L\approx1.55g/L\).

Step4: Calculate the molar mass

From \(
ho=\frac{PM}{RT}\), we can solve for \(M\). \(M=\frac{
ho RT}{P}\). Substitute \(
ho = 2.08g/L\), \(R = 0.0821L\cdot atm/(mol\cdot K)\), \(T = 321K\), \(P = 1.07atm\). \(M=\frac{2.08\times0.0821\times321}{1.07}g/mol\approx51.7g/mol\).

Answer:

Density at \(157^{\circ}C\): \(1.55g/L\); Molar mass: \(51.7g/mol\)