
If V mL of the vapours of substance at NTP weighs W g. The molecular weight of the substance is:
(A)- $\dfrac{W}{V}\times 22400$
(B)-$\dfrac{V}{W}=22400$
(C)- $(W-V)\times 22400$
(D)- $\dfrac{W\times 1}{V\times 22400}$
Answer
578.4k+ views
Hint: The equation of state of a hypothetical ideal gas is given by the ideal gas law. The molecules of hypothetical ideal gas do not attract or repel each other. Ideal gas molecules take up no volume.
Complete answer:
-The empirical relationships between the pressure, the volume, the temperature, and the number of moles of a gas is given by the ideal gas law. It can be used to calculate any of these properties if the other three are known.
-The equation for ideal gas law is as follows-
Ideal gas equation $=PV=nRT$
where R is the gas constant, i.e $R=0.08206\dfrac{L.atm}{K.mol}=8.314\dfrac{J}{K.mol}$
P is the pressure of the gas = 1 atm
V is the volume in litre $=\dfrac{V}{1000}L$
T is the temperature of the gas in Kelvin
N is the amount of substance.
-Let V ml of vapour weigh = W g
22400 ml of vapour will weigh $=\dfrac{W}{V}\times 22400g$
We know that, number of moles of gas $=\dfrac{\text{Given mass }}{\text{Molecular mass}}=\dfrac{\text{W g}}{\text{M g/mol}}$
-Now, the ideal gas law can be written as -
$\begin{align}
& 1atm\times \dfrac{V}{1000}L=\dfrac{W}{M}\times 0.0821\text{ }L\text{ }atm/K\text{ }mol\times 293 \\
& \dfrac{W}{M}\times 24=\dfrac{V}{1000} \\
& \Rightarrow M=\dfrac{W\times 2400}{V}g/mol \\
\end{align}$
So, the correct answer is option A.
Note:
The ideal gas law has some limitations. Since the ideal gas law assumes that gas particles have no volume and are not attracted to each other, if it is so, then on condensing an idea it should be able to condense to a zero volume. But in reality, the real gases do occupy some space, hence the gas law no longer applied here. Another limitation of gas law is that as the kinetic energy decreases while cooling the gas, the particles will eventually move slowly enough that their attractive forces cause them to condense.
Complete answer:
-The empirical relationships between the pressure, the volume, the temperature, and the number of moles of a gas is given by the ideal gas law. It can be used to calculate any of these properties if the other three are known.
-The equation for ideal gas law is as follows-
Ideal gas equation $=PV=nRT$
where R is the gas constant, i.e $R=0.08206\dfrac{L.atm}{K.mol}=8.314\dfrac{J}{K.mol}$
P is the pressure of the gas = 1 atm
V is the volume in litre $=\dfrac{V}{1000}L$
T is the temperature of the gas in Kelvin
N is the amount of substance.
-Let V ml of vapour weigh = W g
22400 ml of vapour will weigh $=\dfrac{W}{V}\times 22400g$
We know that, number of moles of gas $=\dfrac{\text{Given mass }}{\text{Molecular mass}}=\dfrac{\text{W g}}{\text{M g/mol}}$
-Now, the ideal gas law can be written as -
$\begin{align}
& 1atm\times \dfrac{V}{1000}L=\dfrac{W}{M}\times 0.0821\text{ }L\text{ }atm/K\text{ }mol\times 293 \\
& \dfrac{W}{M}\times 24=\dfrac{V}{1000} \\
& \Rightarrow M=\dfrac{W\times 2400}{V}g/mol \\
\end{align}$
So, the correct answer is option A.
Note:
The ideal gas law has some limitations. Since the ideal gas law assumes that gas particles have no volume and are not attracted to each other, if it is so, then on condensing an idea it should be able to condense to a zero volume. But in reality, the real gases do occupy some space, hence the gas law no longer applied here. Another limitation of gas law is that as the kinetic energy decreases while cooling the gas, the particles will eventually move slowly enough that their attractive forces cause them to condense.
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