Question

Density of carbon monoxide is maximum at:
A. 0.5 atm and 273 K
B. 4 atm and 500 K
C. 2 atm and 600 K
D. 6 atm and 1092 K

Answer 1

To answer the density of carbon monoxide, we should use the Ideal gas law. From there will transform the formula to find density, using that density is equal to mass by volume.

Complete step by step solution :
For finding the maximum density of carbon monoxide, we should first use the ideal gas law. Ideal gas law is given by:
PV=nRT
In the above equation P, V and T are the pressure, volume and temperature; n is the amount of substance; and R is the ideal gas constant. R is the same for all gases.
Now, we know that chemical amount (n) (in moles) is equal to total mass of the gas (m) (in kilograms) divided by the molar mass (M) (in kilograms per mole):

& Moles=\frac{total\,mass\,of\,gas}{molar\,mass}=\frac{m}{M} \\\ & \\\ \end{aligned}$$ PV=nRT So, in the above equation we can replace n with$$\frac{m}{M}$$. And now we can write this equation as: $$\begin{aligned} & PV=\frac{m}{M}RT \\\ & P=\frac{m}{V}\frac{RT}{M} \\\ & P=\rho \frac{R}{M}T \\\ & \rho =\frac{PM}{RT} \\\ \end{aligned}$$ Now, R and molar mass is constant. The formula now can be written as: $$\rho =\frac{P}{T}$$ Now, in the options we have given the value of pressure and temperature we will now put in the above equation. 0.5 atm and 273 K $$\rho =\frac{P}{T}$$ $$\rho =\frac{P}{T}=\frac{0.5}{273}=0.0018$$ 4 atm and 500 K $$\rho =\frac{P}{T}=\frac{4}{500}=0.008$$ 2 atm and 600 K $$\rho =\frac{P}{T}=\frac{2}{600}=0.0033$$ 6 atm and 1092 K $$\rho =\frac{P}{T}=\frac{6}{1092}=0.0054$$ From the above calculation, we can say that option B is the correct option. Density of carbon monoxide will be higher at 4 atm pressure and 500 K temperature. **Note** : Carbon monoxide is often called silent killer because it gives no clear warning to its victims. If exposed to low levels of the gas for a long time, it may cause chest pain, dizziness and poor vision.