Pressure, Temperature, and RMS Speed MCQ Quiz - Objective Question with Answer for Pressure, Temperature, and RMS Speed - Download Free PDF

CONCEPT:

Dalton’s law of partial pressures:

• For a mixture of non-reactive ideal gases, the total pressure gets contribution from each gas in the mixture.

\(⇒ P=\frac{1}{3}[n_1m_1\overline{v_1^2}+n_2m_2\overline{v_2^2}+...+n_nm_n\overline{v_n^2}]\)

• In equilibrium, the average kinetic energy of the molecules of different gases will be equal.

\(⇒ \frac{1}{2}m_1\overline{v_1^2}=\frac{1}{2}m_2\overline{v_2^2}=...=\frac{1}{2}m_n\overline{v_n^2}=k_bT\)

So,

⇒ P = (n1 + n2 + ... + nn)kBT

• The mean of the square speed is given as,

\(⇒ \overline{v^2}=\frac{3k_BT}{m}\)

• The square root of \(\overline{v^2}\) is known as root mean square (RMS) speed and is denoted by vrms.
• At the same temperature, lighter molecules have a greater RMS speed.

EXPLANATION:

• We know that for an ideal gas the root mean square (RMS) speed is given as,

\(⇒ v_{rms}=\sqrt{\frac{3k_BT}{m}}\)     -----(1)

Where kB = Boltzmann constant, m = mass of the molecule and T = absolute temperature

• If M and T are the molar mass and the absolute temperature of a gas, then the RMS speed of a gas molecule is given as,

\(⇒ v_{rms}=\sqrt{\frac{3RT}{M}}\)

• Hence, option 3 is correct.

Additional Information

Kinetic interpretation of temperature:

• We know that pressure P of an ideal gas is given as,

\(⇒ P=\frac{1}{3}nm\overline{v^2}\)

• If the volume of the gas is V, then,

\(⇒ PV=\frac{1}{3}nVm\overline{v^2}\)

∵ N = nV

\(\therefore PV=\frac{1}{3}Nm\overline{v^2}\)

• The internal energy of an ideal gas is purely kinetic.
• So the total internal energy E of an ideal gas is given as,

\(⇒ E=N\times\frac{1}{2}m\overline{v^2}\)

• So we can say,

\(⇒ PV=\frac{2}{3}E\)

Where n = number of molecules per unit volume, m = mass of the molecule, N = total number of molecules, and \(\overline{v^2}\) = mean of the squared speed

• If the absolute temperature of an ideal gas is T, then the total internal energy is given as,

\(⇒ E=\frac{3}{2}k_BNT\)

Where kB = Boltzmann constant

• So the average kinetic energy of a molecule is given as,

\(⇒ \frac{E}{N}=\frac{1}{2}m\overline{v^2}=\frac{3}{2}k_BT\)

• The average kinetic energy of a molecule is proportional to the absolute temperature of the gas.
• The average kinetic energy of a molecule is independent of pressure, volume, or the nature of the ideal gas.
• So we can say that the internal energy of an ideal gas depends only on temperature, not on pressure or volume.

\(\displaystyle u = \sqrt {\dfrac {3P}{d}}\)

\(\displaystyle 3180 \: m/s = \sqrt {\dfrac {3P}{8.99 \times 10^{-2} kg/m^3}}\)

\(\displaystyle P = 303034.9 \: N/m^2\)

\(\displaystyle P = \dfrac {303034.9 \: N/m^2}{1.01 \times 10^5 \: N/m^2}\)

\(\displaystyle P = 3.0 \: atm\)

Hence, the pressure on the hydrogen gas is 3.0 atm.

The root mean square (rms) velocity of gas molecules is given by the formula:

v_rms = √(3kT / m), where k is the Boltzmann constant, T is the temperature, and m is the mass of a gas molecule.

From the formula, it is clear that the rms velocity is inversely proportional to the square root of the mass of the gas molecule.

∴ The rms velocity is proportional to 1/√(m). Option 4) is correct.

Calculation:

For a gas, the root mean square (rms) velocity is proportional to the square root of the temperature in Kelvin:

v_rms ∝ √T

v_rms2 / v_rms1 = √(T₂ / T₁)

Let the initial temperature be T1 = 27°C = 27 + 273 = 300 K, and the final temperature be T2 = 90°C = 90 + 273 = 363 K.

The change in rms velocity can be calculated by the ratio of the square roots of the final and initial temperatures:

v_rms2 / v_rms1 = √(363 / 300) ≈ √1.21 ≈ 1.1

% Increase = (1.1 - 1) × 100 = 10%

The V_rms is proportional to the square root of temperature (T) and inversely proportional to the square root of pressure (P).

The initial V_rms is 120 m/s at 88°C (361 K) and 1 atm pressure. We want to find the V_rms at 127°C (400 K) and 2 atm pressure. Using the proportion:

\(V_{\text{rms}} \propto \sqrt{\frac{T}{P}}\)

By substituting the values, the final V_rms is calculated to be approximately 126.3 m/s.

CONCEPT: 

• Gas Molecules are in Constant Motion.
• Root mean Square Speed: The root mean square speed of Gas average of roots of the speed of all Gas molecules present in the given volume.

It is given as 

\(V_{rms} = \sqrt{\frac{3RT}{M}}\)

R = Gas Constant

T = Absolute Temperature of Gas 

M = Molecular Weight of Gas

CALCULATION:

Given V_^rms = 300m/s. 

Let, initial Absolute Temperature is T and initial Molecular weight be M. So,

\(V_{rms} = \sqrt{\frac{3RT}{M}} = 300m/s\) .......  (1)

Now, the Absolute Temperature is halved, So the new absolute Temperature T' = T/2

Also, Molecular weight is doubled, So the new molecular weight M' = 2M

The New root mean square speed

 \(V'_{rms} = \sqrt{\frac{3RT'}{M'}}\).................. (2)

Putting the values of T' and M' in Equation (2) we get

\(V'_{rms} = \sqrt{\frac{3RT}{4M}}\)  

⇒ \(V'_{rms} = \frac{1}{2}\sqrt{\frac{3RT}{M}}\)............  (3)

Comparing Equation (3) and Equation (1)

\(V'_{rms} = \frac{V_{rms}}{2}\)

⇒ V'rms = 300 m/s / 2 = 150 m/s

So, 150 m/s is the Answer.

So, Option 2 is correct.

Additional Information

• The Average of all speeds of the Gas molecule is given as

\(V_{av} = \sqrt{\frac{8RT}{\pi M}}\)

• The Speed passed by the maximum fraction of the total number of molecules, or most probable speed is given as:

\((V_{mp} = \sqrt{\frac{2RT}{M}})\)

CONCEPT:

• Root Mean Square Speed is defined as the square root of the mean of squares of the speed of different molecules.
  • The root-mean-square speed takes into account both molecular weight and temperature, two factors that directly affect the kinetic energy of a material.
  • The rms speed of any homogeneous gas sample is given by:

\(V_{rms}= \sqrt{ {\frac{{3RT}}{M}} }\)

Where R = universal gas constant, T = temperature and M = Molecular mass

CALCULATION:

Given - Initial rms velocity (Vrms1) = V, initial temperature (T1) = 27° C = 300 K and final temperature (T2) = 327° C = 600 K

• As the sample is the same, therefore the molecular mass will be the same. Hence, 

⇒ Vrms ∝ \(\sqrt{T}\)

\(⇒ \frac{V_{rms1}}{V_{rms2}}=\sqrt{{\frac{T_1}{T_2}}}\)

\(⇒ \frac{V}{V_{rms2}}=\sqrt{{\frac{300}{600}}}= \frac{1}{\sqrt{{2}}}\)

⇒ V_rms2 = V\(\sqrt{2}\)

CONCEPT:

RMS Velocity (Vrms):

• The effective velocity of gas particles in a gaseous sample is called Root-mean-square speed (RMS speed).
• RMS velocity is given by:

\({V_{rms}} = \sqrt{\frac{{3\;R\;T}}{M}} \)

Where R = Universal gas constant, T = temperature, and M = Molar mass of the gas.

CALCULATION:

Given - MO₂ = 32 and MH₂ = 2

• RMS velocity of 02 gas is given by:

\(\Rightarrow {V_{O_2}} = \sqrt{\frac{{3\;R\;T}}{M_{O_2}}} =\sqrt{\frac{{3\;R\;T}}{32}}\)    ------- (1)

• RMS velocity of H2 gas is given by:

\(\Rightarrow {V_{H_2}} = \sqrt{\frac{{3\;R\;T}}{M_{H_2}}} =\sqrt{\frac{{3\;R\;T}}{2}}\)  ------- (2)

On dividing equation 1 and 2, we get

\(\Rightarrow \frac{V_{O_2}}{V_{H_2}}= \frac{\sqrt{\frac{{3\;R\;T}}{32}}}{\sqrt{\frac{{3\;R\;T}}{2}}}=\sqrt{\frac{2}{32}}=\frac{1}{4}\)

CONCEPT:

• RMS Velocity (V_rms): The effective velocity of gas particles in a gaseous sample is called Root-mean-square speed (RMS speed).

RMS velocity is given by:

\({V_{rms}} = √{\frac{{3RT}}{M}} \)

Where R is Universal gas constant, T is temperature and M is Molar mass of the gas.

CALCULATION:

CONCEPT:

Root mean square velocity of the gas

• Root mean square velocity (RMS value)is the square root of the mean of squares of the velocity of individual gas molecules.
• It is given as,

​\(⇒ v_{rms}=\sqrt{\frac{3RT}{M}}\)

Where vrms= Root-mean-square velocity, M = Molar mass of the gas (Kg/mole), R= Molar gas constant and T = Temperature in Kelvin

CALCULATION:

• Since the same gas is filled in two containers of the same volume, same temperature, and with the pressure of ratio 1 : 2.

Given M₁ = M₂ = M, V₁ = V₂ = V, T₁ = T₂ = T and \(\frac{P_{1}}{P_{2}}=\frac{1}{2}\)

• The rms speed of the gas is given as,

\(⇒ v_{rms}=\sqrt{\frac{3RT}{M}}\)     -----(1)

• The rms speed for container 1 is given as,

\(⇒ v_{rms1}=\sqrt{\frac{3RT}{M}}\)     -----(2)

• The rms speed for container 2 is given as,

\(\Rightarrow v_{rms2}=\sqrt{\frac{3RT}{M}}\)     -----(3)

By equation 2 and equation 3,

\(\Rightarrow\frac{v_{rms1}}{v_{rms2}}=\frac{1}{1}\)

Hence, option 4 is correct.

CONCEPT:

• Pressure by gas: It is caused by the force exerted by gas molecules colliding with the surfaces of objects.
  • Any surface of an appreciable area experiences a large number of collisions by the gas on the wall, which can result in high pressure.
• Root mean square (r.m.s.) speed: The root-mean-square speed of molecules is the speed at which all the molecules have the same total kinetic energy as in case of their actual speed.

The relation between Root mean square (r.m.s.) speed and pressure are:

\(v_{rms}=\sqrt{\frac{3P}{ρ}}\)

where v_rms is the root mean square speed, P is the pressure and ρ is the density of the gas.

CALCULATION:

Given that P = 6 × 105 Pa; ρ = 5 kg/m3;

\(v_{rms}=\sqrt{\frac{3P}{ρ}}=\sqrt{\frac{3\times6\times10^5}{5}}\)

vrms = 600 m/s

So the correct answer is option 2.

CONCEPT:

The pressure exerted by a gas:

• According to kinetic theory, the molecules of a gas are in a state of continuous random motion.
  • They collide with one another and also with the walls of the vessel.
  • Whenever a molecule collides with the wall, it returns with a changed momentum, and an equal momentum is transferred to the wall (conservation of momentum).
• According to Newton's second law of motion, the rate of transfer of momentum to the wall is equal to the force exerted on the wall.
  • Since a large number of molecules collide with the wall, a steady force is exerted on the wall.
  • The force exerted per unit area of the wall is the pressure of the gas.
  • Hence a gas exerts pressure due to the continuous collisions of its molecules with the walls of the vessel.

EXPLANATION:

• From kinetic theory of gases, the pressure P exerted by an ideal gas of density ρ and rms velocity of its gas molecules C is given by

\(P = \frac{1}{3}ρ v_{rms}^2 - - - - - - - - \left( 1 \right)\)

• Mean kinetic energy of translation per unit volume of the gas is

\(E = \frac{1}{2}ρ v_{rms}^2 - - - - - - - \left( 2 \right) \)

On dividing equations (1) and (2), we get

\(\Rightarrow P=\frac {2}{3}E\)

The above equation can be written as

\(\Rightarrow E=\frac {3}{2}P\)

CONCEPT:

• Since the gas particles are moving in all directions, So the average velocity is zero.
• This is the reason why we use the RMS velocity instead of the average.
• Root mean square (r.m.s.) speed: The root-mean-square speed of molecules is the speed at which all the molecules have the same total kinetic energy as in the case of their actual speed

RMS velocity is calculated by:

\(v = \sqrt{3RT \over M}\)

where T is the temperature of the gas, M is the mass of the gas, and R is the universal gas constant.​

CALCULATION:

Given that v = 500 m/s; Later the temperature is 4 times.

T' = 4T

initially \(v = \sqrt{3RT \over M}\)

Later \(v' = \sqrt{3RT' \over M'}\)

\(v' = \sqrt{3R(4T) \over M}\)

\(v' =2\times \sqrt{3RT \over M}\)

\(v' =2 \times v=2v\)

So the correct answer is option 4.

CONCEPT:

• RMS Velocity (Vrms): The effective velocity of gas particles in a gaseous sample is called Root-mean-square speed (RMS speed).

RMS velocity is given by:

\({V_{rms}} = √ {\frac{{3\;R\;T}}{M}} \)

Where R is Universal gas constant, T is temperature and M is Molar mass of the gas.

CALCULATION:

CONCEPT:

Dalton’s law of partial pressures:

• For a mixture of non-reactive ideal gases, the total pressure gets contribution from each gas in the mixture.

\(⇒ P=\frac{1}{3}[n_1m_1\overline{v_1^2}+n_2m_2\overline{v_2^2}+...+n_nm_n\overline{v_n^2}]\)

• In equilibrium, the average kinetic energy of the molecules of different gases will be equal.

\(⇒ \frac{1}{2}m_1\overline{v_1^2}=\frac{1}{2}m_2\overline{v_2^2}=...=\frac{1}{2}m_n\overline{v_n^2}=k_bT\)

So,

⇒ P = (n1 + n2 + ... + nn)kBT

• The mean of the square speed is given as,

\(⇒ \overline{v^2}=\frac{3k_BT}{m}\)

• The square root of \(\overline{v^2}\) is known as root mean square (RMS) speed and is denoted by vrms.
• At the same temperature, lighter molecules have a greater RMS speed.

EXPLANATION:

• We know that for an ideal gas the root mean square (RMS) speed is given as,

\(⇒ v_{rms}=\sqrt{\frac{3k_BT}{m}}\)

\(⇒ v_{rms}\propto\sqrt{\frac{1}{m}}\)     ----(1)

Where kB = Boltzmann constant, m = mass of the molecule and T = absolute temperature

• By equation 1 we can say that for an ideal gas at the same temperature, lighter molecules have a greater RMS speed.
• Since the mass of the molecule of gas A is more than gas B, therefore the root mean square speed of the molecule of gas A will be less than gas B. Hence, option 2 is correct.

Additional Information

Kinetic interpretation of temperature:

• We know that pressure P of an ideal gas is given as,

\(⇒ P=\frac{1}{3}nm\overline{v^2}\)

• If the volume of the gas is V, then,

\(⇒ PV=\frac{1}{3}nVm\overline{v^2}\)

∵ N = nV

\(\therefore PV=\frac{1}{3}Nm\overline{v^2}\)

• The internal energy of an ideal gas is purely kinetic.
• So the total internal energy E of an ideal gas is given as,

\(⇒ E=N\times\frac{1}{2}m\overline{v^2}\)

• So we can say,

\(⇒ PV=\frac{2}{3}E\)

Where n = number of molecules per unit volume, m = mass of the molecule, N = total number of molecules, and \(\overline{v^2}\) = mean of the squared speed

• If the absolute temperature of an ideal gas is T, then the total internal energy is given as,

\(⇒ E=\frac{3}{2}k_BNT\)

Where kB = Boltzmann constant

• So the average kinetic energy of a molecule is given as,

\(⇒ \frac{E}{N}=\frac{1}{2}m\overline{v^2}=\frac{3}{2}k_BT\)

• The average kinetic energy of a molecule is proportional to the absolute temperature of the gas.
• The average kinetic energy of a molecule is independent of pressure, volume, or the nature of the ideal gas.
• So we can say that the internal energy of an ideal gas depends only on temperature, not on pressure or volume.