Question

One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown in the figure. Calculate the heat rejected by the gas in the path CA and the heat absorbed by the gas in the path AB.

• A. $\frac{7}{2}{P}_0{V}_0$.
• B. $\frac{8}{2}{P}_0{V}_0$.
• C. $\frac{3}{2}{P}_0{V}_0$.
• D. $\frac{5}{2}{P}_0{V}_0$.

Answer 1

The correct option is D $\frac{5}{2}{P}_0{V}_0$.

Given :

solution :

$Q_{AB}=n{C}_V\Delta T$

$\Rightarrow n\left(\frac{3}{2}R\right)(T_B-T_A)$

$\Rightarrow \frac{3}{2}(p_B{V}_B-p_A{V}_A)$

$\Rightarrow 3p_0{V}_0$

$Q_{CA}=n{C}_p\delta T$

$\Rightarrow n\left(\frac{5}{2}R\right)(T_A-T_C)$

$\Rightarrow \frac{5}{2}(p_A{V}_A-p_C{V}_C)$

$\Rightarrow \frac{5}{2}(p_0{V}_0-2p_0{V}_0)=-\frac{5}{2}{p}_0{V}_0$

hence for the b part question answer is $\frac{5}{2}{p}_0{V}_0$

hence the correct opt : D