Question

An ideal monatomic gas is confined in a cylinder by a spring loaded piston of cross-section $8\times {10}^{-3}{m}^2$.Initially the gas is at 300K and occupies a volume of $2.4\times {10}^{-3}{m}^3$ and the spring is in its relaxed position. The gas is heated by a small electric heater until the piston moves out slowly by 0.1m. Calculate (a) the final temperature of the gas and (b) the heat supplied (in J) by the heater. The force constant of the spring is 8000 N/m and atmospheric pressure $1\times {10}^{5}N/m^2$.

• A. 400 K , 200 J
• B. 300K , 100J
• C. 300 K , 350 J
• D. 800 K , 120J

Answer 1

The correct option is D 800 K , 120J

As here initially
$P_I=P_0=1\times {10}^{5}N/m^2,$
$V_I=V_0=2.4\times {10}^{-3}{m}^3,$
$T_I=T_0=300K$
and finally
$P_F=P_0+\frac{Kx}{A}=1\times {10}^5+\frac{8000\times 0.1}{8\times {10}^{-3}}$
$=2\times {10}^{5}N/m^2$,
$V_F=V_0+Ax=2.4\times 10+0.1\times 8\times {10}^{-3}$
$=3.2\times {10}^{-3}{m}^3,$
So from gas equation PV = $\mu$RT, i.e.
$\frac{P_I{V}_I}{T_I}=\frac{P_F{V}_F}{T_F}$

we have
$T_F=\frac{P_F}{P_I}\times \frac{V_F}{V_I}\times T_I$
$=\frac{2\times {10}^5}{1\times {10}^5}\times \frac{3.2\times {10}^{-3}}{2.4\times {10}^{-3}}\times 300=800K$
$\left(b\right)As\Delta W=\int PdV,$
$here,P=P_0+\left(\frac{kx}{A}\right)$
$so\Delta W=P+\frac{kx}{A}dx$
$=(P_{0}A+kx)dx$
$i.e.=P_{0}Ax+\frac{1}{2}k{x}^2$
$=[{10}^5\times 8\times {10}^{-3}\times 0.1+\frac{1}{2}\times 8000\times (0.1{)}^2]$
$i.e.\Delta W=80+40=120J$