Solveeit Logo
Login

Question

Question: A \[0.5d{m^3}\] flask contains gas \(A\) and \(1d{m^3}\) flask contains gas \(B\) at the same temper...

A 0.5dm30.5d{m^3} flask contains gas AA and 1dm31d{m^3} flask contains gas BB at the same temperature. If density of AA is twice that of Band the molar mass of AA if half of BB, then the ratio of pressure exerted by gases is?
A)PAPB=3A)\dfrac{{{P^A}}}{{{P^B}}} = 3
B)PAPB=4B)\dfrac{{{P^A}}}{{{P^B}}} = 4
C)PAPB=2C)\dfrac{{{P^A}}}{{{P^B}}} = 2
D)PAPB=1D)\dfrac{{{P^A}}}{{{P^B}}} = 1

Explanation

Solution

The ideal gas equation is the relationship among the volume, the temperature, the pressing factor, and the measure of a gas can be combined into the ideal gas law, PV = nRT.

Complete step by step solution:
We know that the ideal gas equation, is
=PV=nRTPV = nRT.
Where PP= pressure,
VV= Volume,
nn= amount of substance,
RR= Ideal gas constant,
TT= temperature.
We can write n=Mass(m)Molarmass(M)n = \dfrac{{Mass(m)}}{{Molarmass(M)}}.
Now, put the value of nn in the ideal gas equation.
=PV=mMRTPV = \dfrac{m}{M}RT.
=P=mVRTMP = \dfrac{m}{V}\dfrac{{RT}}{M}.
We know that density(d)=mM(d) = \dfrac{m}{M} .
SO, P=dRTMP = d\dfrac{{RT}}{M}.
For AA the equation will be
=dA=PAMARTA{d_A} = \dfrac{{{P_A}{M_A}}}{{R{T_A}}}
And for BB the equation will be
=dB=PBMBRTB{d_B} = \dfrac{{{P_B}{M_B}}}{{R{T_B}}}.
So, we get
=PAMARTA=PBMBRTB\dfrac{{{P_A}{M_A}}}{{R{T_A}}} = \dfrac{{{P_B}{M_B}}}{{R{T_B}}}.
We are given that,
VA=0.5dm3,VB=1dm3 TA=TB=T dA=2dB MA=12MB  {V_A} = 0.5d{m^3},{V_B} = 1d{m^3} \\\ {T_A} = {T_{\mathbf{B}}} = T \\\ {d_A} = 2{d_B} \\\ {M_A} = \dfrac{1}{2}{M_B} \\\
So, we can write
=PAMARTA=2(PBMBRTB)\dfrac{{{P_A}{M_A}}}{{R{T_A}}} = 2\left( {\dfrac{{{P_B}{M_B}}}{{R{T_B}}}} \right).
=PAMART=2(PB×2MART)\dfrac{{{P_A}{M_A}}}{{RT}} = 2\left( {\dfrac{{{P_B} \times 2{M_A}}}{{RT}}} \right).
=PA=4PB{P_A} = 4{P_B}
=PAPB=4\dfrac{{{P_A}}}{{{P_B}}} = 4.

So, the correct option is BB.

Additional Information:
The kinetic theory assumptions about ideal gases are comprised of atoms which are inconsistent arbitrary movement in straight lines, The atoms carry on as unbending circles, Pressing factor is because of impacts between the atoms and the dividers of the compartment. All collisions, both between the actual particles and between the atoms and the dividers of the compartment, are totally flexible, The temperature of the gas is corresponding to the normal dynamic energy of the particles.

Note:
Remember in reality, there is nothing of the sort as an ideal gas, yet an ideal gas is a valuable reasonable model that permits us to see how gases react to changing conditions. As we shall see, under many conditions, most ideal gases exhibit behavior that closely approximated that of the ideal gas. Therefore, it helps us to predict the behavior of ideal gases under given conditions.