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Question: At \(0^\circ C\) the value of density of fixed mass of an ideal gas divided by its pressure is x. At...

At 0∘C0^\circ C the value of density of fixed mass of an ideal gas divided by its pressure is x. At
100∘C100^\circ C, this quotient is :
A. 100273x\dfrac{{100}}{{273}}x
B. 273100x\dfrac{{273}}{{100}}x
C. 273373x\dfrac{{273}}{{373}}x
D. 373273x\dfrac{{373}}{{273}}x

Explanation

Solution

Hint-
We can solve this problem using the ideal gas equation .The ideal gas equation given as
PV=nRTPV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, T is the temperature.
By substituting for volume as mass divided by density we can get the ratio of density and pressure at both temperatures. On comparing these ratios we can arrive at the final answer.

Step by step solution:
It is given that the value of density of fixed mass of an ideal gas divided by its pressure is x at 0∘C0^\circ C.
We need to find the value at 100∘C100^\circ C.
We know that for an ideal gas we have the ideal gas equation given as
PV=nRTPV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, T is the temperature.
let the pressure at 0∘C0^\circ C be P1{P_1}
We know that density is mass divided by volume. Hence we can write the volume as mass divided by density.
that is,
V=MρV = \dfrac{M}{\rho }
At 0∘C0^\circ Cthe temperature T=0+273 K=273 KT = 0 + 273\,K = 273\,K
Therefore at 0∘C0^\circ C we can write the ideal gas equation as ,
P1Mρ1=nR×273{P_1}\dfrac{M}{{{\rho _1}}} = nR \times 273
⇒ρ1P1=nMR×273\Rightarrow \dfrac{{{\rho _1}}}{{{P_1}}} = \dfrac{{nM}}{{R \times 273}}.............(1)
The value of this ratio is given as x. Therefore we can write,
ρ1P1=nMR×273=x\dfrac{{{\rho _1}}}{{{P_1}}} = \dfrac{{nM}}{{R \times 273}} = x
Let the pressure at 100∘C100^\circ C be P2{P_2} . the temperature T2{T_2} will be
T2=100+273=373 K{T_2} = 100 + 273 = 373\,K
The ideal gas equation at this temperature can be written as ,
ρ2P2=nMR×373\dfrac{{{\rho _2}}}{{{P_2}}} = \dfrac{{nM}}{{R \times 373}}................(2)
Now let us divide equation 1 by equation 2.
Then we will get ,
ρ1P1ρ2P2=nMR×273nMR×373\dfrac{{\dfrac{{{\rho _1}}}{{{P_1}}}}}{{\dfrac{{{\rho _2}}}{{{P_2}}}}} = \dfrac{{\dfrac{{nM}}{{R \times 273}}}}{{\dfrac{{nM}}{{R \times 373}}}}
Therefore , we get
ρ1P1ρ2P2=373273\dfrac{{\dfrac{{{\rho _1}}}{{{P_1}}}}}{{\dfrac{{{\rho _2}}}{{{P_2}}}}} = \dfrac{{373}}{{273}}
Now we have ρ1P1=x\dfrac{{{\rho _1}}}{{{P_1}}} = x
On substituting this and rearranging the equation we get,
xρ2P2=373273\dfrac{x}{{\dfrac{{{\rho _2}}}{{{P_2}}}}} = \dfrac{{373}}{{273}}
ρ2P2=273373x\dfrac{{{\rho _2}}}{{{P_2}}} = \dfrac{{273}}{{373}}x
This is the value of density of fixed mass of an ideal gas divided by its pressure at 100∘C100^\circ C .

So the correct answer is option C.

Note: Here we were able to cancel out the mass since it is given it is fixed. The number of moles n that appears in the ideal gas equation also cancels out because it is the ratio of given mass by molar mass of the substance. Since mass remains fixed, the number of moles at both temperatures will also be equal.