Question

An ideal gas expands isothermally performing 4.30 x 10³ of work in the process. Calculate (a)

the change in internal energy of the gas, and (b) the heat absorbed during this expansion.​

Answer 1

The solution of the given query is explained below in the explanation portion.

Step-by-step explanation:

Given value is:

Work,

⇒
`W=4.30* 10^3 \ J`

(a)

The change in the internal energy will be:

⇒
`\Delta V=(3)/(2)nR(\Delta T)`

Throughout the isothermal procedure, The temperature will be constant "ΔT = 0".

then,

⇒
`\Delta V = (3)/(2)nR(0)`

⇒
`\Delta V = 0 \ J`

(b)

As we know,

⇒
`\Delta V = Q-W`

On substituting the value, we get

⇒
`0=Q-W`

On adding W both sides, we get

⇒
`W=Q-W+W`

⇒
`W=Q`

On substituting the given value of "W", we get

⇒
`Q=4.30* 10^3 \ J`