Question

w = \int_{1}^{2} p dv
t : constant
ideal gas ->
pr = nrt
p = \frac{nrt}{v}
w = \int_{1}^{2} \frac{nrt}{v} dv
w = nrt \int_{1}^{2} \frac{dv}{v}
w = nrt \ln v_{1}^{2}=nrt \ln v_{i}^{f}=nrt \ln|\frac{v_{f}}{v_{i}}|=nrt \ln|\frac{p_{i}}{p_{f}}|
t : const. or isothermal
pr = nrt
p_{1}v_{1}=p_{2}v_{2}
p_{3}v_{3}=p_{4}v_{4}
\frac{v_{f}}{v_{i}} = \frac{p_{i}}{p_{f}}
dt = 0
\delta e = 0
\varphi=\delta e + w
\varphi = w

Explanation:

Step1: Start with work formula

Given $W=\int_{1}^{2}P dV$ for an ideal - gas process at constant temperature ($T$ is constant). For an ideal gas, $PV = nRT$, so $P=\frac{nRT}{V}$. Substituting $P$ into the work formula gives $W=\int_{1}^{2}\frac{nRT}{V}dV$.

Step2: Factor out constants

Since $n$, $R$, and $T$ are constants for this isothermal process, we can factor them out of the integral: $W = nRT\int_{1}^{2}\frac{dV}{V}$.

Step3: Evaluate the integral

The integral of $\frac{1}{V}$ with respect to $V$ is $\ln V$. So, $W=nRT[\ln V]_{1}^{2}=nRT(\ln V_{2}-\ln V_{1})$.

Step4: Use logarithm properties

Using the property of logarithms $\ln a-\ln b=\ln\frac{a}{b}$, we get $W = nRT\ln\frac{V_{2}}{V_{1}}$. Also, from $P_{1}V_{1}=P_{2}V_{2}$ (for an isothermal ideal - gas process), $\frac{V_{2}}{V_{1}}=\frac{P_{1}}{P_{2}}$, so $W = nRT\ln\frac{P_{1}}{P_{2}}$.

Answer:

$W = nRT\ln\frac{V_{2}}{V_{1}}=nRT\ln\frac{P_{1}}{P_{2}}$