Question

A sample of an ideal gas undergoes an isothermal process as shown by the curve $ AB $ in the $ pV $ diagram. If $ \Delta Q $ , $ \Delta\,U $ and $ \Delta\,W $ represent the amount of heat absorbed the change in internal energy and the work done respectively, then which of the following statement is correct?

• A. $ \Delta Q=+\, ve, \Delta U= 0, \Delta W=-ve $
• B. $ \Delta Q=+\, ve, \Delta U= 0, \Delta W=+ve $
• C. $ \Delta Q=+\, ve, \Delta U= 0, \Delta W=0 $
• D. $ \Delta Q=+\, ve, \Delta U= +ve, \Delta W=+ve $

Answer 1

Answer: (B)

From the graph of an ideal gas given in the question, the process is an isothermal expansion because volume of the ideal gas increases
Given, $V_{B} >\,V_{A}$ and $p_{B}<\,P_{A}$
$\because \Delta T=0$ (isothermal process)
$\therefore $ Change in internal energy
$\Delta U=0 \left(\because\Delta U \propto\Delta T\right)$
Since we know, in isothermal process
$\Delta Q=\Delta W=n\cdot RT \ln \left(\frac{V_{f}}{V_{p}}\right)$
or $\Delta Q=\Delta W=nRT \ln \left(\frac{V_{B}}{V_{A}}\right)$
Thus, $ \Delta Q=+ve$ and $\Delta w=+ve$
because $ V_{B} >\, V_{A}$ (from the graph)