Question

A container is filled with an ideal diatomic gas to a pressure and volume of P₁ and V₁, respectively. The gas is then warmed in a two-step process that increases the pressure by a factor of three and the volume by a factor of two. Determine the amount of energy transferred to the gas by heat if the first step is carried out at constant volume and the second step at constant pressure. (Use any variable or symbol stated above as necessary.)

Q=?

Answer 1

Solution:

Given,

Initial values, P₁ and V₁, Using ideal gas equation

T₁ = P₁V₁/nR .

Step 1 : Constant volume, Pressure is increased to 3P₁

T₂ = 3P₁V₁/nR .

The heat energy added in this step, C_v = 5/2R for diatomic

Q₁ = nC_vT

= n[5/2]R*[3P₁V₁ - P₁V₁]/nR

= 5P₁V₁.

Step 2: Constant pressure, Volume increased to 2V₁

T₃ = 3P₁(2V₁)/nR = 6P₁V₁/nR

The heat energy added in this step C_p = 7/2R for diatomic,

Q₂ = nC_pT

= n[7/2]R[6P₁V₁ - 3P₁V₁]/nR

= 10.5P₁V₁.

the total Heat added is,

Q_net = Q₁ + Q₂

= 15.5P₁V₁.

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