During the derivation for the work done in isothermal reversible expansion of an ideal gas, the following expression appears.
$d W = (P - d P) d V = P d V + d P \cdot d V$

True

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False

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Solution

The correct option is B False
Derivation of work done in isothermal reversible expansion
$d W = P \times A \times d l = P \times d V$
The amount of work done by isothermal reversible expansion of an ideal gas from $V_{1}$ to $V_{2}$ is
$W = \int_{V_{1}}^{V_{2}} P d V$
From ideal gas equation, $P = \frac{n R T}{V}$
Hence $W = \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V = n R T \int_{V_{1}}^{V_{2}} \frac{d V}{V}$
On integration we get
$W = n R T ln \frac{V_{2}}{V_{1}}$
Since
$P_{1} V_{1} = P_{2} V_{2} \frac{P_{1}}{P_{2}} = \frac{V_{2}}{V_{1}} W = n R T ln \frac{P_{1}}{P_{2}}$

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During the derivation for the work done in isothermal reversible expansion of an ideal gas, the following expression appears.dW=(P−dP)dV=PdV+dP⋅dV

During the derivation for the work done in isothermal reversible expansion of an ideal gas, the following expression appears.
$d W = (P - d P) d V = P d V + d P \cdot d V$