Question

The density of ${{O}_{2}}$ is 16 ${kg}/{{{m}^{3}}}\;$ at NTP. At what temperature its density will be 14 ${kg}/{{{m}^{3}}}\;$? Consider that the pressure remains constant.
    (A) 39${}^\circ C$
    (B) 43${}^\circ C$
    (C) 57${}^\circ C$
    (D) None of the above

Answer 1

Hint: From ideal gas law we can derive the relationship between density and temperature. Then by rearranging the equation we can find the equation for the new temperature. On substituting all the given values in the question and on conversion from kelvin to degree Celsius we will get the new temperature at which density would be 14 ${kg}/{{{m}^{3}}}\;$.

Complete answer:
  As we know, Ideal gas equation can be written as,
PV=nRT
  Where P is the pressure of ideal gas
             V is the volume of ideal gas
             n is the amount of substance of gas (number of moles)
             R is the universal gas constant
             T is the temperature of ideal gas
In the given question the relation between temperature and density must be found. Therefore, an alternate form of this law can be used. The chemical amount (n) (in moles) is equal to the total mass of the gas (m) (in kilograms) divided by the molar mass (M) (in kilograms per mole) and thus we can rewrite the ideal gas equation as
    $PV=\quad \dfrac{m}{M}RT$
By rearranging the equation, we get,
    $P=\quad \dfrac{m}{V}\dfrac{RT}{M}$
Since $\dfrac{m}{V}$ is density (ρ), the equation can be rewritten as
   $P=\dfrac{\rho }{M}RT$
From this relation, we can derive the relation between temperature and density as (pressure is kept constant)
  ${{\rho }_{1}}{{T}_{1}}={{\rho }_{2}}{{T}_{2}}$
We are given the values of ${{T}_{1}}$=293 K (Since given condition is in NTP)
                                              ${{\rho }_{1}}$= 16 ${kg}/{{{m}^{3}}}\;$
                                              ${{\rho }_{2}}$= 14 ${kg}/{{{m}^{3}}}\;$
                                              ${{T}_{2}}$ = ?
By substituting the values in the above equation, we get,
   16×293=14× ${{T}_{2}}$
   ${{T}_{2}}$= $\dfrac{16\times 293}{14}$
        = 334 K
For converting ${{T}_{2}}$ from kelvin to ${}^\circ C$;
    ${{T}_{2}}$ = 334−273
         = 61${}^\circ C$
Therefore 61${}^\circ C$ would be the temperature at which density will be 14 ${kg}/{{{m}^{3}}}\;$.

So, the answer is option(D) None of the above.

Note: We should be careful while choosing ${{T}_{1}}$. The given condition is NTP (Normal Temperature and Pressure). So, the temperature is 293 K whereas for STP (Standard Temperature and Pressure) the temperature is 273 K.