
Two non-reactive monatomic ideal gases have their atomic masses in the ratio 2:3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4:3. The ratio of their densities is:
A. 1:4
B. 1:2
C. 6:9
D. 8:9
Answer
567.9k+ views
Hint: In kinetic theory gases, we assume several data regarding gases to make their behavior close to that of ideal gas in order to analyze their behavior properly and reducing the calculation part. An ideal gas is assumed to have no interaction with the corresponding particles of gases.
Formula used:
$\dfrac{P_1M_1}{\rho_1} = \dfrac{P_2M_2}{\rho_2}$
Complete answer:
Partial pressure of a gas is the pressure exerted by individual gas in a container having more than one gases. We are given the ratio of molecular masses as 2:3 i.e.
$\dfrac{M_1}{M_2} = \dfrac23$and the ratio of their partial pressure as 4:3 i.e. $\dfrac {P_1}{P_2} = \dfrac43$.
Thus, using the equation of state $\dfrac{P_1M_1}{\rho_1} = \dfrac{P_2M_2}{\rho_2}$, we have;
$\dfrac{\rho_2}{\rho_1} = \dfrac{P_2}{P_1}\times \dfrac{M_2}{M_1} =\dfrac34\times\dfrac32 = \dfrac98$
So, $\dfrac{\rho_1}{\rho_2} = \dfrac 89$.
So, the correct answer is “Option D”.
Additional Information:
Equation of ideal gas – The most fundamental equation of an ideal gas is $PV=nRT$. Here, ‘P’ is the partial pressure of a gas contained in a chamber of volume ‘V’ having temperature ‘T’ and quantity of ‘n’ moles. Here ‘R’ is the universal gas constant and is equal to $R = 8.314 JK^{-1}mol^{-1}$. But for getting the equation in terms of density and molecular mass, we put $n=\dfrac{m}{M}$. Putting it in $PV=nRT$, we get:
$PV = \dfrac mM RT \implies PM = \dfrac mV RT$
And since $\dfrac mV = \rho = density\ of\ gas$, thus
We get the final equation as $PM = \rho RT$, where ‘M’ is the molecular mass of gas.
Note:
Many times, we need the relation between different parameters of a gas like, temperature, pressure, volume, moles, density, etc. All these parameters could be found using the basic ideal gas equation. Just we need little modification and we can get the desired results. Students are advised to practice as much as problems based on the ideal gas equation.
Formula used:
$\dfrac{P_1M_1}{\rho_1} = \dfrac{P_2M_2}{\rho_2}$
Complete answer:
Partial pressure of a gas is the pressure exerted by individual gas in a container having more than one gases. We are given the ratio of molecular masses as 2:3 i.e.
$\dfrac{M_1}{M_2} = \dfrac23$and the ratio of their partial pressure as 4:3 i.e. $\dfrac {P_1}{P_2} = \dfrac43$.
Thus, using the equation of state $\dfrac{P_1M_1}{\rho_1} = \dfrac{P_2M_2}{\rho_2}$, we have;
$\dfrac{\rho_2}{\rho_1} = \dfrac{P_2}{P_1}\times \dfrac{M_2}{M_1} =\dfrac34\times\dfrac32 = \dfrac98$
So, $\dfrac{\rho_1}{\rho_2} = \dfrac 89$.
So, the correct answer is “Option D”.
Additional Information:
Equation of ideal gas – The most fundamental equation of an ideal gas is $PV=nRT$. Here, ‘P’ is the partial pressure of a gas contained in a chamber of volume ‘V’ having temperature ‘T’ and quantity of ‘n’ moles. Here ‘R’ is the universal gas constant and is equal to $R = 8.314 JK^{-1}mol^{-1}$. But for getting the equation in terms of density and molecular mass, we put $n=\dfrac{m}{M}$. Putting it in $PV=nRT$, we get:
$PV = \dfrac mM RT \implies PM = \dfrac mV RT$
And since $\dfrac mV = \rho = density\ of\ gas$, thus
We get the final equation as $PM = \rho RT$, where ‘M’ is the molecular mass of gas.
Note:
Many times, we need the relation between different parameters of a gas like, temperature, pressure, volume, moles, density, etc. All these parameters could be found using the basic ideal gas equation. Just we need little modification and we can get the desired results. Students are advised to practice as much as problems based on the ideal gas equation.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

