question_answer

A gas at pressure \[6\times {{10}^{s}}N/{{m}^{2}}\]and volume \[1{{m}^{3}}\]expands to \[3\text{ }{{m}^{3}}\] and its pressure falls to\[4\times {{10}^{5}}N/{{m}^{2}}.\] Given, that the indicator diagram is a          straight line, work done by the system is

A)                  \[6\times {{10}^{5}}J\]                                 

B)  \[3\times {{10}^{5}}J\]

C)                  \[4\times {{10}^{5}}J\]                                 

D)  \[10\times {{10}^{5}}J\]

Correct Answer: D

Solution :

                The indicator diagram can be shown as below                                 Work done by the system                 = area under p-V diagram                 = area of rectangle BCDE + area of \[\Delta \]ABC                 \[=4\times 10\times 2\times \frac{2\times {{10}^{5}}\times 2}{2}\]                 \[=8\times {{10}^{5}}\times 2\times {{10}^{5}}\]                 \[=10\times {{10}^{5}}J\]