Ideal Gas equation MCQ Quiz - Objective Question with Answer for Ideal Gas equation - Download Free PDF
Last updated on May 4, 2025
Latest Ideal Gas equation MCQ Objective Questions
Ideal Gas equation Question 1:
One mole of an ideal gas is subjected to a two step reversible process \(A-B\) and \(B-C\). The pressure at \(A\) and \(C\) is same. Mark the correct option(s):
Answer (Detailed Solution Below)
Ideal Gas equation Question 1 Detailed Solution
CONCEPT:
Ideal Gas Law and Volume Relationship
- The ideal gas law is given by:
PV = nRT
- For an ideal gas undergoing a reversible process, the volume and temperature are related by:
\(\frac{V_C}{V_A} = \frac{T_C}{T_A}\)
- As pressure is the same at points A and C, the ratio of volumes depends on the ratio of temperatures.
EXPLANATION:
- Volume at Point B:
- Using the ideal gas equation at point B:
\(V_B = \frac{nRT_B}{P}\)
- Given data:
n = 1 , R = 0.082 , T_B = 600K , P = 3 atm
- Calculation:
\(V_B = \frac{1 \times 0.082 \times 600}{3} = 16.42 liters\)
- Using the ideal gas equation at point B:
- Volume Relationship Between C and A:
- Since pressure is constant at A and C:
\(\frac{V_C}{V_A} = \frac{T_C}{T_A} = \frac{3}{1} \\ V_C = 3 \times V_A\)
- Since pressure is constant at A and C:
Correct Options: The correct options are Option 3: Volume of gas at C = 3 × volume of gas at A & Option 4: Volume of gas at B is 16.42 liters
Ideal Gas equation Question 2:
A gaseous organic compound has a density of \( 2.5\,kg/m^3 \) at \( 2 \,atm \) and at \( 273^{\circ}\,C \) . The molecular formula of the compound can be :
Answer (Detailed Solution Below)
Ideal Gas equation Question 2 Detailed Solution
\( d = \dfrac{PM}{RT} \)
\( \therefore \, 2.5 = \dfrac{2 \times M_{gas}}{0.082\times 546} \)
\( \therefore \, M_{gas} = 56 \)
Options C and D are correct.
Ideal Gas equation Question 3:
Two moles of an ideal monoatomic gas undergoes a cyclic process ABCA as shown in V-T diagram below
Work done by the gas during the entire cycle is:
Answer (Detailed Solution Below)
Ideal Gas equation Question 3 Detailed Solution
CONCEPT:
Work Done in a Cyclic Process
W = nR * (Tf - Ti) / (K - 1)
- In a cyclic process, the gas goes through a series of transformations that eventually bring it back to its original state. The total work done by the gas is the area enclosed by the path on the PV diagram or can be calculated from the individual segments of the cycle in the V-T diagram.
- For an ideal gas undergoing a polytropic process, where VT = constant, the work done in each segment can be calculated using the formula:
- Tf and Ti are the final and initial temperatures for the process.
- R is the ideal gas constant, n is the number of moles of the gas, and K is the polytropic index (which is 2 in this case for the given process).
EXPLANATION:
WBC = nR * (Tf - Ti) / (K - 1)
WBC = 2 * R * (1200 - 600) / (2 - 1) = 1200 * R
-1200R ln 2 + 1200R + 0 = 1200R (1 - ln 2)
- From the diagram, the process AB and BC are part of the cyclic process, with the condition VT = constant for the segment BC. This indicates that the gas undergoes a polytropic process between points B and C, where the work done can be calculated using the formula mentioned above.
- For segment BC, the work done is given by:
- Now, the total work done by the gas during the entire cycle is the sum of the work done during each segment:
- WAB + WBC + WCA
- Therefore, the work done during the entire cycle is 1200R (1 - ln 2).
Hence, the correct answer is option 3) 1200R(1 - ln 2).
Ideal Gas equation Question 4:
The density of an ideal gas can be given by __________, where P, V, M, T and R respectively denote pressure, volume, molar-mass, temperature and universal gas constant.
Answer (Detailed Solution Below)
Ideal Gas equation Question 4 Detailed Solution
Concept:
The density \(\rho\) of an ideal gas is its mass per unit volume. By relating the ideal gas law to the definition of density, we can derive a formula for the density of an ideal gas in terms of its pressure P, molar mass M, temperature T, and the universal gas constant R.
Explanation:
Ideal Gas Law: The ideal gas law is given by:
PV = 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
- Moles to Mass:
The number of moles n can be expressed in terms of mass m and molar mass M:
\(n = \frac{m}{M}\)
- Substitute n in the Ideal Gas Law:
By substituting \(n = \frac{m}{M}\) into the ideal gas equation,
we get: \(PV = \frac{m}{M} RT\)
Rearranging terms to solve for density \(\rho\) mass per unit volume \(\rho = \frac{m}{V}\)
\(m = \rho V PV = \frac{\rho V}{M} RT\)
Simplifying the equation:
\(P = \frac{\rho RT}{M}\)
Solve for Density:
To find the expression for density \(\rho\), solve for \(\rho\):
\(\rho = \frac{PM}{RT}\)
Conclusion:
The density of an ideal gas is given by:\(\rho = \frac{PM}{RT}\)
Ideal Gas equation Question 5:
A gas in a vessel is heated in such a way that its pressure and volume both becomes two times. The temperature of the gas expressed in kelvin scale becomes __________ the initial temperature.
Answer (Detailed Solution Below)
Ideal Gas equation Question 5 Detailed Solution
Concept:
Ideal Gas or Perfect Gas:
- If there are no intermolecular attractive forces and all collisions between atoms or molecules are entirely elastic, then a substance is said to be an ideal gas.
- Every change in internal energy causes a change in temperature since all of the internal energy in such gas is kinetic energy.
- Hydrogen, oxygen, nitrogen, and noble gases such as helium, and neon all are ideal gases.
Explanation:
Given Data:
Pressure and volume both become double as the gaseous vessel is heated.
Assuming that,
Before heating:
\(P\) is the initial pressure of the gas
\(V\) is the initial volume of the gas
\(T\) is the initial temperature of the gas
After heating
\(P^{'}\) is the final pressure that becomes \(2P\)
\(V^{'}\)is the final volume that becomes \(2V\)
\(T^{'}\) is the final temperature.
Now, applying the ideal gas equation:
\(PV\;=\;nRT\)
\(\dfrac{PV}{T}\;=\;\dfrac{P^{'}V^{^{'}}}{T^{'}}\)
Also, the pressure and volume become twice as the vessel is heated. So,
\(\dfrac{T^{'}}{T}\;=\;\dfrac{2P2V}{PV}\)
\(\dfrac{T^{'}}{T}\;=\;4\)
\(\mathbf{T^{'}\;=\;4T}\)
Conclusion:
Thus, the temperature of the gas expressed in the kelvin scale becomes four times the initial temperature.
Additional Information
Ideal Gas Equation:
The equation of state for a fictitious ideal gas is known as the ideal gas law. Although it has several drawbacks, it is a good approximation of the behavior of numerous gases under various circumstances. The formula for the ideal gas equation is :
\(\mathbf{PV\;=\;nRT} \)
where P represents the pressure of a gas
V is the volume,
n is the moles,
R is the Gas constant, and
T represents the temperature.
This relation derives from the combination of Charles's law, Boyles's law, and Avogadro's law:
As stated by Boyle's Law,
The volume and the pressure a gas exerts have an inverse relationship at constant n and T.
\(V\;\propto\;\dfrac{1}{P}\)
Charles' Law states that
The volume of a gas has a direct relationship with temperature when p and n are constant.
\(V\;\propto\;T\)
As stated by Avogadro's Law,
The volume of a gas has a direct relationship with the number of moles of gas when p and T are constant.
\(V\;\propto\;n\)
A combination of all three laws makes an equation:
\(V\;\propto\;\dfrac{nT}{P}\)
OR
\(\mathbf{PV\;=\;nRT} \)
Top Ideal Gas equation MCQ Objective Questions
Ideal Gas equation Question 6:
The density of an ideal gas can be given by __________, where P, V, M, T and R respectively denote pressure, volume, molar-mass, temperature and universal gas constant.
Answer (Detailed Solution Below)
Ideal Gas equation Question 6 Detailed Solution
Concept:
The density \(\rho\) of an ideal gas is its mass per unit volume. By relating the ideal gas law to the definition of density, we can derive a formula for the density of an ideal gas in terms of its pressure P, molar mass M, temperature T, and the universal gas constant R.
Explanation:
Ideal Gas Law: The ideal gas law is given by:
PV = 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
- Moles to Mass:
The number of moles n can be expressed in terms of mass m and molar mass M:
\(n = \frac{m}{M}\)
- Substitute n in the Ideal Gas Law:
By substituting \(n = \frac{m}{M}\) into the ideal gas equation,
we get: \(PV = \frac{m}{M} RT\)
Rearranging terms to solve for density \(\rho\) mass per unit volume \(\rho = \frac{m}{V}\)
\(m = \rho V PV = \frac{\rho V}{M} RT\)
Simplifying the equation:
\(P = \frac{\rho RT}{M}\)
Solve for Density:
To find the expression for density \(\rho\), solve for \(\rho\):
\(\rho = \frac{PM}{RT}\)
Conclusion:
The density of an ideal gas is given by:\(\rho = \frac{PM}{RT}\)
Ideal Gas equation Question 7:
A gas in a vessel is heated in such a way that its pressure and volume both becomes two times. The temperature of the gas expressed in kelvin scale becomes __________ the initial temperature.
Answer (Detailed Solution Below)
Ideal Gas equation Question 7 Detailed Solution
Concept:
Ideal Gas or Perfect Gas:
- If there are no intermolecular attractive forces and all collisions between atoms or molecules are entirely elastic, then a substance is said to be an ideal gas.
- Every change in internal energy causes a change in temperature since all of the internal energy in such gas is kinetic energy.
- Hydrogen, oxygen, nitrogen, and noble gases such as helium, and neon all are ideal gases.
Explanation:
Given Data:
Pressure and volume both become double as the gaseous vessel is heated.
Assuming that,
Before heating:
\(P\) is the initial pressure of the gas
\(V\) is the initial volume of the gas
\(T\) is the initial temperature of the gas
After heating
\(P^{'}\) is the final pressure that becomes \(2P\)
\(V^{'}\)is the final volume that becomes \(2V\)
\(T^{'}\) is the final temperature.
Now, applying the ideal gas equation:
\(PV\;=\;nRT\)
\(\dfrac{PV}{T}\;=\;\dfrac{P^{'}V^{^{'}}}{T^{'}}\)
Also, the pressure and volume become twice as the vessel is heated. So,
\(\dfrac{T^{'}}{T}\;=\;\dfrac{2P2V}{PV}\)
\(\dfrac{T^{'}}{T}\;=\;4\)
\(\mathbf{T^{'}\;=\;4T}\)
Conclusion:
Thus, the temperature of the gas expressed in the kelvin scale becomes four times the initial temperature.
Additional Information
Ideal Gas Equation:
The equation of state for a fictitious ideal gas is known as the ideal gas law. Although it has several drawbacks, it is a good approximation of the behavior of numerous gases under various circumstances. The formula for the ideal gas equation is :
\(\mathbf{PV\;=\;nRT} \)
where P represents the pressure of a gas
V is the volume,
n is the moles,
R is the Gas constant, and
T represents the temperature.
This relation derives from the combination of Charles's law, Boyles's law, and Avogadro's law:
As stated by Boyle's Law,
The volume and the pressure a gas exerts have an inverse relationship at constant n and T.
\(V\;\propto\;\dfrac{1}{P}\)
Charles' Law states that
The volume of a gas has a direct relationship with temperature when p and n are constant.
\(V\;\propto\;T\)
As stated by Avogadro's Law,
The volume of a gas has a direct relationship with the number of moles of gas when p and T are constant.
\(V\;\propto\;n\)
A combination of all three laws makes an equation:
\(V\;\propto\;\dfrac{nT}{P}\)
OR
\(\mathbf{PV\;=\;nRT} \)
Ideal Gas equation Question 8:
For radial probability curves, which of the following is/are correct?
Answer (Detailed Solution Below)
Ideal Gas equation Question 8 Detailed Solution
\(N = n-l-1\)
Option A is correct as per given above image.
Option B is correct as the number of radial or spherical nodes is given by \(n-1\).
Option C is correct.
Option D is wrong. It has 2 angular nodes.
Hence, Option A,B,C are correct.
Ideal Gas equation Question 9:
Which of the following curve is correct for an ideal gas?
Answer (Detailed Solution Below)
Ideal Gas equation Question 9 Detailed Solution
The curves which are correct for an ideal gas are curve "C" & "D" because in this P is directly proportional to n and V is directly proportional to T.
Hence, Options "C" & "D" are correct answers.
Ideal Gas equation Question 10:
Which of the following curve is correct for an ideal gas?
Answer (Detailed Solution Below)
Ideal Gas equation Question 10 Detailed Solution
The curves which are correct for an ideal gas are curve "C" & "D" because in this P is directly proportional to n and V is directly proportional to T.
Hence, Options "C" & "D" are correct answers.
Ideal Gas equation Question 11:
One mole of an ideal gas is subjected to a two step reversible process \(A-B\) and \(B-C\). The pressure at \(A\) and \(C\) is same. Mark the correct option(s):
Answer (Detailed Solution Below)
Ideal Gas equation Question 11 Detailed Solution
CONCEPT:
Ideal Gas Law and Volume Relationship
- The ideal gas law is given by:
PV = nRT
- For an ideal gas undergoing a reversible process, the volume and temperature are related by:
\(\frac{V_C}{V_A} = \frac{T_C}{T_A}\)
- As pressure is the same at points A and C, the ratio of volumes depends on the ratio of temperatures.
EXPLANATION:
- Volume at Point B:
- Using the ideal gas equation at point B:
\(V_B = \frac{nRT_B}{P}\)
- Given data:
n = 1 , R = 0.082 , T_B = 600K , P = 3 atm
- Calculation:
\(V_B = \frac{1 \times 0.082 \times 600}{3} = 16.42 liters\)
- Using the ideal gas equation at point B:
- Volume Relationship Between C and A:
- Since pressure is constant at A and C:
\(\frac{V_C}{V_A} = \frac{T_C}{T_A} = \frac{3}{1} \\ V_C = 3 \times V_A\)
- Since pressure is constant at A and C:
Correct Options: The correct options are Option 3: Volume of gas at C = 3 × volume of gas at A & Option 4: Volume of gas at B is 16.42 liters
Ideal Gas equation Question 12:
A gaseous organic compound has a density of \( 2.5\,kg/m^3 \) at \( 2 \,atm \) and at \( 273^{\circ}\,C \) . The molecular formula of the compound can be :
Answer (Detailed Solution Below)
Ideal Gas equation Question 12 Detailed Solution
\( d = \dfrac{PM}{RT} \)
\( \therefore \, 2.5 = \dfrac{2 \times M_{gas}}{0.082\times 546} \)
\( \therefore \, M_{gas} = 56 \)
Options C and D are correct.
Ideal Gas equation Question 13:
Two moles of an ideal monoatomic gas undergoes a cyclic process ABCA as shown in V-T diagram below
Work done by the gas during the entire cycle is:
Answer (Detailed Solution Below)
Ideal Gas equation Question 13 Detailed Solution
CONCEPT:
Work Done in a Cyclic Process
W = nR * (Tf - Ti) / (K - 1)
- In a cyclic process, the gas goes through a series of transformations that eventually bring it back to its original state. The total work done by the gas is the area enclosed by the path on the PV diagram or can be calculated from the individual segments of the cycle in the V-T diagram.
- For an ideal gas undergoing a polytropic process, where VT = constant, the work done in each segment can be calculated using the formula:
- Tf and Ti are the final and initial temperatures for the process.
- R is the ideal gas constant, n is the number of moles of the gas, and K is the polytropic index (which is 2 in this case for the given process).
EXPLANATION:
WBC = nR * (Tf - Ti) / (K - 1)
WBC = 2 * R * (1200 - 600) / (2 - 1) = 1200 * R
-1200R ln 2 + 1200R + 0 = 1200R (1 - ln 2)
- From the diagram, the process AB and BC are part of the cyclic process, with the condition VT = constant for the segment BC. This indicates that the gas undergoes a polytropic process between points B and C, where the work done can be calculated using the formula mentioned above.
- For segment BC, the work done is given by:
- Now, the total work done by the gas during the entire cycle is the sum of the work done during each segment:
- WAB + WBC + WCA
- Therefore, the work done during the entire cycle is 1200R (1 - ln 2).
Hence, the correct answer is option 3) 1200R(1 - ln 2).
Ideal Gas equation Question 14:
A gas in a vessel is heated in such a way that its pressure and volume both becomes two times. The temperature of the gas expressed in kelvin scale becomes __________ the initial temperature.
Answer (Detailed Solution Below)
Ideal Gas equation Question 14 Detailed Solution
Concept:
Ideal Gas or Perfect Gas:
- If there are no intermolecular attractive forces and all collisions between atoms or molecules are entirely elastic, then a substance is said to be an ideal gas.
- Every change in internal energy causes a change in temperature since all of the internal energy in such gas is kinetic energy.
- Hydrogen, oxygen, nitrogen, and noble gases such as helium, and neon all are ideal gases.
Explanation:
Given Data:
Pressure and volume both become double as the gaseous vessel is heated.
Assuming that,
Before heating:
\(P\) is the initial pressure of the gas
\(V\) is the initial volume of the gas
\(T\) is the initial temperature of the gas
After heating
\(P^{'}\) is the final pressure that becomes \(2P\)
\(V^{'}\)is the final volume that becomes \(2V\)
\(T^{'}\) is the final temperature.
Now, applying the ideal gas equation:
\(PV\;=\;nRT\)
\(\dfrac{PV}{T}\;=\;\dfrac{P^{'}V^{^{'}}}{T^{'}}\)
Also, the pressure and volume become twice as the vessel is heated. So,
\(\dfrac{T^{'}}{T}\;=\;\dfrac{2P2V}{PV}\)
\(\dfrac{T^{'}}{T}\;=\;4\)
\(\mathbf{T^{'}\;=\;4T}\)
Conclusion:
Thus, the temperature of the gas expressed in the kelvin scale becomes four times the initial temperature.
Additional Information
Ideal Gas Equation:
The equation of state for a fictitious ideal gas is known as the ideal gas law. Although it has several drawbacks, it is a good approximation of the behavior of numerous gases under various circumstances. The formula for the ideal gas equation is :
\(\mathbf{PV\;=\;nRT} \)
where P represents the pressure of a gas
V is the volume,
n is the moles,
R is the Gas constant, and
T represents the temperature.
This relation derives from the combination of Charles's law, Boyles's law, and Avogadro's law:
As stated by Boyle's Law,
The volume and the pressure a gas exerts have an inverse relationship at constant n and T.
\(V\;\propto\;\dfrac{1}{P}\)
Charles' Law states that
The volume of a gas has a direct relationship with temperature when p and n are constant.
\(V\;\propto\;T\)
As stated by Avogadro's Law,
The volume of a gas has a direct relationship with the number of moles of gas when p and T are constant.
\(V\;\propto\;n\)
A combination of all three laws makes an equation:
\(V\;\propto\;\dfrac{nT}{P}\)
OR
\(\mathbf{PV\;=\;nRT} \)