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If 1.00 mol of an ideal monatomic gas initially at 77 K absorbs 100 J of thermal energy, what is the final temperature? Assume the volume does not change.

Respuesta :

Answer:

85 K

Explanation:

T₀ = initial temperature of the gas = 77 K

T = final temperature of the gas = ?

n = number of moles of monoatomic gas = 1.00 mol

R = universal gas constant = 8.314

c = heat capacity at constant volume = (1.5) R = (1.5) (8.314) = 12.5

Q = Amount of heat absorbed = 100 J

Amount of heat absorbed is given as

Q = n c (T - T₀)

100 = (1) (12.5) (T - 77)

T = 85 K

onyebuchinnaji

The final temperature of the ideal monatomic gas is 85.02 K.

Quantity of heat

The quantity of heat absorbed by the monatomic gas is given by the following formula;

Q = ncΔT

Q = nc(T₂ - T₁)

Where;

  • n is the number of moles of the gas
  • T₁ is initial temperature of the gas
  • T₂ is the final temperature of the gas
  • c is heat capacity at of the gas at constant volume

Q = n(1.5R)(T₂ - T₁)

where;

  • R is gas constant = 8.314

100 = 1(1.5 x 8.314)(T₂ - 77)

100 = 12.47(T₂ - 77)

T₂ - 77 = 100/12.47

T₂ - 77 = 8.012

T₂  = 8.012 + 77

T₂  = 85.02 K

Thus, the final temperature of the ideal monatomic gas is 85.02 K.

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