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

Answer:

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

Explanation:

Given value is:

Work,

⇒  $W=4.30\times 10^3 \ J$

(a)

The change in the internal energy will be:

⇒  $\Delta V=\frac{3}{2}nR(\Delta T)$

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

then,

⇒  $\Delta V = \frac{3}{2}nR(0)$

⇒  $\Delta V = 0 \ J$

(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

⇒     $Q=4.30\times 10^3 \ J$