Suppose the internal energy of an ideal gas rises by 2.89 103 J at a constant pressure of 1.00 105 Pa, while the system gains 6.44 103 J of energy by heat. Find the change in volume of the system.

Respuesta :

Answer:

[tex]\Delta V=35.5\ L[/tex]

Explanation:

According to the first law of thermodynamics:-

[tex]\Delta U = q - w[/tex]

Where,  

U is the internal energy  = [tex]2.89\times 10^3\ J[/tex]

q is the heat  = [tex]6.44\times 10^3\ J[/tex] (heat is gained)

w is the work done  = ?

[tex]\Delta U = q - w[/tex]

[tex]w=q-\Delta U = 6.44\times 10^3\ J-2.89\times 10^3\ J= 3.55\times 10^3\ J[/tex]

The expression for the calculation of work done is shown below as:

[tex]w=P\times \Delta V[/tex]

Where, P is the pressure,  [tex]1.00\times 10^5\ Pa[/tex]

[tex]\Delta V[/tex] is the change in volume

Also, 1 J = 1000 PaL

So,

[tex]w=3.55\times 10^6\ PaL[/tex]

From the question,  

[tex]3.55\times 10^6=1.00\times 10^5\times \Delta V[/tex]

[tex]\Delta V=35.5\ L[/tex]