Question

An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cross section $8.0\times {10}^{-3}{m}^2$. 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 state as shown in figure. The gas is heated by a small heater until the piston moves out slowly by $0.1m$. The force constant of the spring is 8000 N/m and the atmospheric pressure is $1\times {10}^{5}N/m^2$. The cylinder and the piston are thermally insulated. The piston and the spring are massless and there is no friction between the piston and the cylinder. The final temperature of the gas will be :

(Neglect the heat loss through the lead wires of the heater. The heat capacity of the heater coil is also negligible)

• A. $800K$
• B. $1000K$
• C. $500K$
• D. $300K$

Answer 1

The correct option is A $800K$

$F=kx=8000\times 0.1\times 800N$

Pressure exerted on the piston by the spring is
$\Delta P=\frac{F}{A}=\frac{800}{8\times {10}^{-3}}=1\times {10}^{5}N/m^2$

The total pressure $P_2$ of the gas inside cylinder is .
$P_2=P_{atm}+\Delta P=1\times {10}^5+1\times {10}^5=2\times {10}^{5}N/m^2$

Since the piston has moved outwards, there has been an increase of $\Delta V$ in the volume of the gas, i.e.,

$\Delta V=A\times x=(8\times {10}^{-3})\times \left(0.1\right)$

$=8\times {10}^{-4}{m}^3$

The final volume of the gas

$V_2=V_1+\Delta V=2.4\times {10}^{-3}+8\times {10}^{-4}=3.2\times {10}^{-3}{m}^3$

Let $T_2$ be the final temperature of gas. Then

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}\Rightarrow T_2=\frac{P_2{V}_2}{P_1{V}_1}{T}_1$

$=300\times \frac{2\times {10}^5\times 3.2\times {10}^{-3}}{{10}^5\times 2.4\times {10}^{-3}}$

$=800K$