Question

work done by a system at constant temperature isothermal expansion
$varphi = delta e+w$, $w(by)>0$, $w(on)<0$, $varphi (add)>0$, $varphi (lose)<0$
$delta e=\frac{f}{2}nrdelta t$
$delta t = t_{f}-t_{i}$
isothermal : $|delta t = 0|$
$delta e = 0$
$varphi = w$
$w=int dw=int padx$
$=int pdv$
$w=int_{1}^{2} pdv
ightarrow$
$w = fcdotdelta x$
$|w| = fdelta x$
$f = pa$
$p = f/a$
$dw = fdx$
$w_{12}=int_{1}^{2} dw=int_{1}^{2} fdx$
$average$
instantaneous

Explanation:

Step1: Identify the key concept

For isothermal expansion, $\Delta T = 0$. Since internal - energy change $\Delta E=\frac{f}{2}nR\Delta T$, for $\Delta T = 0$, $\Delta E = 0$.

Step2: Recall the first - law of thermodynamics

The first - law of thermodynamics is $\Delta U=\Delta E + W$. Since $\Delta E = 0$ for isothermal process, $\Delta U = W$.

Step3: Express work in integral form

Work done by a gas is $W=\int PdV$. For a quasi - static process, $P = \frac{F}{A}$ and $W=\int Fdx$. Also, $dW = Fdx$ and $W_{12}=\int_{1}^{2}dW=\int_{1}^{2}Fdx$.

Answer:

For an isothermal expansion of a system, $\Delta E = 0$ and work done $W=\int PdV$ where $P$ is the pressure of the gas and $V$ is the volume.