Question

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

• A. $12\times {10}^{5}J$
• B. $10\times {10}^{5}J$
• C. $6\times {10}^{5}J$
• D. $4\times {10}^{5}J$

Answer 1

Answer: (B)

$10\times {10}^{5}J$

As per the given graph we get the work done as area under the PV graph. Thus
$areaofrectangle=(4\times 2)+areaoftriangle(=\frac{1}{2}\times 2\times 2)$
$=10\times {10}^{5}J$