Five moles of an ideal monatomic gas with an initial temperature of 121 ∘c expand and, in the process, absorb an amount of heat equal to 1200 j and do an amount of work equal to 2020 j .
what is the final temperature of the gas?

  T₂ ≈ 107.85∘C

Explanation:

The question didn't state if the volume is constant or not as such, we can apply the first law of thermodynamic

From the first law of thermodynamic,

ΔU =  Q - W

where ΔU = Internal Energy, Q = Quantity of heat absorbed, W = Amount of work done.

Q = 1200 J and W = 2020 J

∴ ΔU = 1200 -2020 = -820 J.

Using the ideal gas equation,

ΔU = 3/2nRΔTequation 1

where n = number of moles, R = Molar gas constant, ΔT = Change in temperature = (T₂ - T₁).

Modifying equation 1,

ΔU = 3/2nR(T₂ -T₁)equation 2.

making T₂ the subject of the relation in equation 2,

T₂ =  {2/3(ΔU)/nR}+T₁ equation 3

where T₁=121∘C, R= 8.314 J / mol, n=5 moles, ΔU=-820 J

Substituting these values into equation 3,

∴ T₂ ={ 2/3(-820)/(5×8.314)}+121

   T₂ = {2×(-820)/ (3×5×8.314)}+121  

   T₂={-1640/124.71}+ 121

   T₂ = {-13.151} + 121

 ∴T₂  = 121 - 13.151 = 107. 849∘C

    T₂ ≈ 107.85∘C

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

electrical power