The absolute temperature of a sample of monatomic ideal gas is doubled at constant volume. what effect

Answer;

Doubles and Remains the same

Explanation;The effect of doubling the absolute temperature of a sample of a monoatomic ideal gas is that,the pressure doubles and density of the sample of gas remains the same.According to ideal gas equation; PV = nRT; Where P is pressure and V is the Volume, n is the number of moles, R is the ideal gas constant and T is the absolute temperature.Therefore, when the temperature of the mono atomic ideal gas is doubled, the pressure of the gas will also doubles.However, in a closed chamber mass of the ideal gas is invariant, since density depends only on the mass and volume therefore the density of the ideal is gas will remain the same.

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

B) erini just took the test and it is correct