Question

A quantity of a monatomic ideal gas undergoes a process in which both its pressure and volume are doubled as shown in the figure above.
DATA:
V₀ = 0.23 m³
P₀ = 14500 Pa.
What is the change of the internal energy of the gas?

Tries 0/20

What was the work done by the gas during the expansion?

Tries 0/20

What amount of heat flowed into the gas during the expansion?

Tries 0/20

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Answer 1

Change in internal energy:

The internal energy of monatomic ideal gas at pressure P and volume V is given by .

Here, P = 2P₀ and V = 2V₀.

Therefore,

  

Now, change in internal energy is,

  

Here, P₀ = 14500 Pa and

V₀ = 0.23 m³

Change in internal energy

  = 15007.5 J

Work done during expansion :

The area under the path on a PV diagram is equal to the magnitude of work done on the gas.

Formula to calculate the area under PV diagram is,

• W₁ is the width of the rectangle A

• H₁ is the height of the rectangle A

• B₁ is the base of the triangle B

• H₂ is the height of the triangle B

Substitute V₀ for W₁, P₀ for H₁,V₀ for B₁ and P₀ for H₂ to find the area of the PV diagram.

  

Since the final volume is higher than the initial volume, the work done on the gas is negative.

Thus, the work done on the gas is,

  

  

= - 5002.5 J

Amount of heat flow during expansion :

Heat flow,  

= 15007.5 -(-5002.5)

= 15007.5 + 5002.5

= 20010 J