Question

The PV graph of a process is a shown. Here $P_0={10}^{5}Pa$ and $V_0={10}^{-3}{m}^3$. The working substance is 2 moles of a mono – atomic gas. The change in internal energy of the gas and heat supplied in the process is (in joule)

• A. 100, 100
• B. 750, 1050
• C. 1000, 2000
• D. 700, 1400

Answer 1

The correct option is B 750, 1050

Applying I law of thermodynamics.
$\begin{array}{cc}{Q}_{AB}=\Delta U_{AB}+W_{AB}& \Delta U_{AB}=n{C}_v\Delta T\\ W_{AB}=\frac{1}{2}(P_0+2P_0)\left(2V_0\right)& =2\times \frac{3}{2}R[T_B-T_A]\\ =3P_0{V}_0& =3R[\frac{6P_0{V}_0}{2R}-\frac{P_0{V}_0}{2R}]\end{array}$
$\Delta U_{AB}=\frac{15\times {10}^5\times {10}^{-3}}{2}$
$=750J$
$Q_{AB}=\frac{15P_0{V}_0}{2}+3P_0{V}_0$
$=\frac{21P_0{V}_0}{2}$
$=\frac{21}{2}\times {10}^5\times {10}^{-3}$
$=1050J$.