Question

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.

Answer 1

`\Delta V=35.5\ L`

Step-by-step explanation:

According to the first law of thermodynamics:-

`\Delta U = q - w`

Where,

U is the internal energy =
`2.89* 10^3\ J`

q is the heat =
`6.44* 10^3\ J` (heat is gained)

w is the work done = ?

`\Delta U = q - w`

`w=q-\Delta U = 6.44* 10^3\ J-2.89* 10^3\ J= 3.55* 10^3\ J`

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

`w=P* \Delta V`

Where, P is the pressure,
`1.00* 10^5\ Pa`

`\Delta V` is the change in volume

Also, 1 J = 1000 PaL

So,

`w=3.55* 10^6\ PaL`

From the question,

`3.55* 10^6=1.00* 10^5* \Delta V`

`\Delta V=35.5\ L`