The ratio of specific heats (γ = Cₚ / Cᵥ) for a monoatomic gas is given by: γ = (f + 2) / f where f is the degrees of freedom. For a monoatomic gas, f = 3 (translational motion only), so γ = (3 + 2) / 3 = 5/3 ≈ 1.67 So, the ratio of specific heats for monoatomic gas is 1.67. Click here for more: httRead more
The ratio of specific heats (γ = Cₚ / Cᵥ) for a monoatomic gas is given by:
γ = (f + 2) / f
where f is the degrees of freedom. For a monoatomic gas, f = 3 (translational motion only), so
γ = (3 + 2) / 3 = 5/3 ≈ 1.67
So, the ratio of specific heats for monoatomic gas is 1.67.
According to the kinetic theory of gases, gas molecules are in constant random motion and exert pressure due to their collisions with the walls of the container. The pressure P is given by: P = (1/3) (N/V) m v̄² where: - N is the number of molecules, - V is the volume, - m is the mass of a molecule,Read more
According to the kinetic theory of gases, gas molecules are in constant random motion and exert pressure due to their collisions with the walls of the container. The pressure P is given by:
P = (1/3) (N/V) m v̄²
where:
– N is the number of molecules,
– V is the volume,
– m is the mass of a molecule,
– v̄² is the mean square velocity.
Every collision transfers momentum to the wall, and the result of many of these collisions is measurable gas pressure.
The root mean square speed of gases molecules is provided by: vₘₛ = √(3RT/M) - Here:, R: It is the Universal Gas Constant, T : This is Absolute temperature M : It refers to molar mass of a gas. Here this formula provides a condition showing that rms increases with an increment in temperature while tRead more
The root mean square speed of gases molecules is provided by:
vₘₛ = √(3RT/M)
– Here:,
R: It is the Universal Gas Constant,
T : This is Absolute temperature
M : It refers to molar mass of a gas.
Here this formula provides a condition showing that rms increases with an increment in temperature while the rms declines as molar mass increases.
The average kinetic energy (KE_avg) of a gas molecule is given by: KE_avg = (3/2) k_B T where: - k_B is Boltzmann's constant, - T is the absolute temperature. Since KE_avg ∝ T, the average kinetic energy of a gas molecule is directly proportional to the temperature in Kelvin. This means that as tempRead more
The average kinetic energy (KE_avg) of a gas molecule is given by:
KE_avg = (3/2) k_B T
where:
– k_B is Boltzmann’s constant,
– T is the absolute temperature.
Since KE_avg ∝ T, the average kinetic energy of a gas molecule is directly proportional to the temperature in Kelvin. This means that as temperature increases, the kinetic energy of gas molecules also increases.
The kinetic energy of one mole of an ideal gas is given by: KE = (3/2) RT where: - R is the universal gas constant, - T is the absolute temperature in Kelvin. This equation follows from the kinetic theory of gases, which states that the total kinetic energy of gas molecules is directly proportionalRead more
The kinetic energy of one mole of an ideal gas is given by:
KE = (3/2) RT
where:
– R is the universal gas constant,
– T is the absolute temperature in Kelvin.
This equation follows from the kinetic theory of gases, which states that the total kinetic energy of gas molecules is directly proportional to temperature.
The ratio of specific heats (Cₚ / Cᵥ) for a monoatomic gas is:
The ratio of specific heats (γ = Cₚ / Cᵥ) for a monoatomic gas is given by: γ = (f + 2) / f where f is the degrees of freedom. For a monoatomic gas, f = 3 (translational motion only), so γ = (3 + 2) / 3 = 5/3 ≈ 1.67 So, the ratio of specific heats for monoatomic gas is 1.67. Click here for more: httRead more
The ratio of specific heats (γ = Cₚ / Cᵥ) for a monoatomic gas is given by:
γ = (f + 2) / f
where f is the degrees of freedom. For a monoatomic gas, f = 3 (translational motion only), so
γ = (3 + 2) / 3 = 5/3 ≈ 1.67
So, the ratio of specific heats for monoatomic gas is 1.67.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-14/
According to the kinetic theory of gases, the pressure exerted by a gas is due to:
According to the kinetic theory of gases, gas molecules are in constant random motion and exert pressure due to their collisions with the walls of the container. The pressure P is given by: P = (1/3) (N/V) m v̄² where: - N is the number of molecules, - V is the volume, - m is the mass of a molecule,Read more
According to the kinetic theory of gases, gas molecules are in constant random motion and exert pressure due to their collisions with the walls of the container. The pressure P is given by:
P = (1/3) (N/V) m v̄²
where:
– N is the number of molecules,
– V is the volume,
– m is the mass of a molecule,
– v̄² is the mean square velocity.
Every collision transfers momentum to the wall, and the result of many of these collisions is measurable gas pressure.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-14/
The root mean square (rms) speed of gas molecules is given by:
The root mean square speed of gases molecules is provided by: vₘₛ = √(3RT/M) - Here:, R: It is the Universal Gas Constant, T : This is Absolute temperature M : It refers to molar mass of a gas. Here this formula provides a condition showing that rms increases with an increment in temperature while tRead more
The root mean square speed of gases molecules is provided by:
vₘₛ = √(3RT/M)
– Here:,
R: It is the Universal Gas Constant,
T : This is Absolute temperature
M : It refers to molar mass of a gas.
Here this formula provides a condition showing that rms increases with an increment in temperature while the rms declines as molar mass increases.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-14/
The average kinetic energy of a gas molecule is directly proportional to:
The average kinetic energy (KE_avg) of a gas molecule is given by: KE_avg = (3/2) k_B T where: - k_B is Boltzmann's constant, - T is the absolute temperature. Since KE_avg ∝ T, the average kinetic energy of a gas molecule is directly proportional to the temperature in Kelvin. This means that as tempRead more
The average kinetic energy (KE_avg) of a gas molecule is given by:
KE_avg = (3/2) k_B T
where:
– k_B is Boltzmann’s constant,
– T is the absolute temperature.
Since KE_avg ∝ T, the average kinetic energy of a gas molecule is directly proportional to the temperature in Kelvin. This means that as temperature increases, the kinetic energy of gas molecules also increases.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-14/
The kinetic energy of one mole of an ideal gas is given by:
The kinetic energy of one mole of an ideal gas is given by: KE = (3/2) RT where: - R is the universal gas constant, - T is the absolute temperature in Kelvin. This equation follows from the kinetic theory of gases, which states that the total kinetic energy of gas molecules is directly proportionalRead more
The kinetic energy of one mole of an ideal gas is given by:
KE = (3/2) RT
where:
– R is the universal gas constant,
– T is the absolute temperature in Kelvin.
This equation follows from the kinetic theory of gases, which states that the total kinetic energy of gas molecules is directly proportional to temperature.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-14/