What is the temperature change of 1.0 mol of a monotomic gas if its thermal energy is increased by 1.0 J ? Express your answer in kelvins.

0.08K

Explanation:

When temperature changes, the corresponding change in thermal energy of a gas is given by:

ΔE (thermal) = 3/2nRΔT

Defining the parameters:

ΔE (thermal) = Increase in thermal energy of the mono atomic gas = 1.0J

n = number of moles of the gas = 1.0mol

R = Ideal gas constant = 8.314J/mol/K

ΔT = change in temperature. This is what we need to find.

Rearranging the equation to make ΔT subject of the formula,

ΔT= 2 x ΔE (thermal) / (3 x n x R)

Therefore, ΔT = 2 x 1.0J / (3 x 1.0mol x 8.314J/mol/K)

ΔT = 2.0J / 24.942J/K

ΔT = 0.0802K

ΔT = 0.08K

The temperature change of 1.0mol of a monoatomic gas if its thermal energy is increased by 1.0J is 0.08K.

490 n

f=125n a=2.5m/s2

the speed of the cart after 3 seconds of low fan speed is ⇒ 54.0 cm/s.

the speed of the cart after 5 seconds of medium fan speed is ⇒ 120.0 cm/s.

the speed of the cart after 2 seconds of high fan speed is ⇒ 64.0 cm/s. 

just did the lab.