Question

Air is contained in a frictionless piston-cylinder arrangement as shown in the figure.

The atmospheric pressure is 100 kPa and the initial pressure of air in the cylinder is 105 kPa. The area of piston is 300$c{m}^2$. Heat is now added and the piston moves slowly from its initial position until it reaches the stops. The spring constant of the linear spring is 12.5 N/mm. Considering air inside the cylinder as the system, the work interaction is J. (round off to the nearest integer).

• A. 544

Answer 1

The correct option is A 544
As per given information

$P_o=100kPa,P_1=105kPa$
$K=12.5N/mm=12.5kN/m$
$A=300c{m}^2=300\times {10}^{-4}{m}^2$
$x=8cm=8\times {10}^{-2}m$
1-2constant pressure
$W_{1-2}=P_1\times A\times x$ $=105\times 300\times {10}^{-4}\times 8\times {10}^{-2}$
$=0.252kJ=252J$
$P^2=105kPa$

$P_3\times A=P_{2}A+Kx$
$P_3=p2+\frac{Kx}{A}$
$=105+\frac{12.5\times 8\times {10}^{-2}}{300\times {10}^{-4}}$
$=138.33kPa$
$w_{2-3}=\frac{1}{2}(P_2+P_3)\times (V_3-V3)$
$=\frac{1}{2}(105+138.33)A\times x$
$=\frac{1}{2}\left(243.33\right)\times 300\times {10}^{-4}\times 8\times {10}^{-2}$
$W_{2-3}=0.2919kJ=291.9J$

$\therefore W_{total}=W_{1-2}+W_{2-3}$
$=0.5439kJ=543.91=544J$
Alternative solution:
Total work = workdone because of 105 kPa pressure + workdone against spring which is equal to energy stored in spring
$Workdone=P_1\times A\times 2x+\frac{1}{2}k.x^2$
$=105\times 300\times {10}^{-4}\times 2\times 8\times {10}^{-2}+\frac{1}{2}\times 12.5\times (8\times {10}^{-2}{)}^2$
$=0.504+0.04$
$=544kJ=544J$