Question

What is the density of acetylene gas measured at pressure of $ 0.75 $ atm and temperature of $ {250^ \circ }C $ ?

Answer 1

Hint: For calculating the mass, volume, pressure and temperature of a gaseous sample, a number of laws are used. Boyles and Charles law gives individually the relationship between the volume of a gaseous sample with its pressure and temperature.

Complete Step By Step Answer:
When we combine Boyle's law and Charles law together it will lead to formation of another law which is commonly known as combined gas law. Combined gas law gives the simultaneous effects of variation of pressure and temperature on the volume of gaseous samples.
The mathematical equation used to express combined gas law is commonly known as gas equation.
Combined gas law is: $ PV = nRT $
Where $ P = $ pressure of gas
 $ V = $ Volume of gas
 $ n = $ Moles of gas
 $ R = $ Gas constant having value $ 0.0821 $ $ Latm/Kmol $
 $ T = $ Temperature of gas in kelvin
From the question we have a value of pressure of $ 0.75 $ atm and temperature of $ {250^ \circ }C $ .
First, we have to convert the temperature of gas from Celsius to kelvin by adding $ 273.5 $ to the original value.
Temperature of acetylene $ = 250 + 273.5 $ $ {\rm K} $
After solving, we get
Temperature of acetylene $ = 523.5 $ $ {\rm K} $
Now put these values in gas equation to obtain the moles of acetylene present
 $ PV = nRT $
 $ n = \dfrac{{PV}}{{RT}} $
We know that mole fraction of gas can be expressed in terms of its molar mass and weight of the sample.
Mole of gas $ \left( n \right) $ $ = \dfrac{m}{M} $
Where $ m = $ mass of the gas
 $ M = $ Molecular mass of the gas
Now put the value of $ n $ , in gas equation
 $ \dfrac{m}{M} = \dfrac{{PV}}{{RT}} $
On rearranging the equation, we get
 $ \dfrac{m}{V} = \dfrac{{PM}}{{RT}} $
We already know that the ratio of mass to volume is known as density. Hence, we can replace the value of $ \left( {\dfrac{m}{V}} \right) $ by density $ \left( \rho \right) $ .
Now the gas equation in terms of density will become-
 $ \rho = \dfrac{{PM}}{{RT}} $
Now put all the values in this equation to determine the density of acetylene gas
 $ P = 0.75 $ atm
 $ T = 523.5 $ $ {\rm K} $
 $ R = 0.0821 $ $ Latm/Kmol $
 $ M = $ Molecular mass of acetylene $ \left( {{C_2}{H_2}} \right) $
 $ M = 2 \times 12 + 2 \times 1 $
On solving, we get
 $ M = 26 $
Now calculate density by using gas equation
 $ \rho = \dfrac{{0.75 \times 26}}{{0.0821 \times 523.5}} $
On solving the above equation, we get
 $ \rho = 0.45 $ $ g/L $
 $ \Rightarrow $ The density of acetylene gas measured at pressure of $ 0.75 $ atm and temperature of $ {250^ \circ }C $ is $ 0.45 $ $ g/L $ .

Note:
Value of gas constant changes when we change the unit of $ R $ . For example, the value of gas constant is $ 82.1 $ when it is expressed in $ mL atm/Kmol $ . Gas constant represents the total work done which occurs at per degree per mole of the gas. Do not forget to convert the unit of temperature in kelvin.