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 a relaxed state. The gas is heated by a small heater coil $H$. The force constant of the spring is $8000N/m$ and the atmospheric pressure is $1.0\times {10}^{5}Pa$. The cylinder and the piston are thermally insulated. The piston is massless and there is no friction between the piston and the cylinder. Neglect heat loss through lead wires of the heater. The heat capacity of the heater coil is negligible. With all the above assumptions, if the gas is heated by the heater until the piston moves out slowly by $0.1m$, then the final temperature is

• A. $400K$
• B. $800K$
• C. $1200K$
• D. $300K$

Answer 1

The correct option is B $800K$

Initially, the pressure of the gas in the cylinder is atmospheric pressure as spring is in relaxed state. Therefore,
$P_1$- atmospheric pressure = $1\times {10}^{5}N/m^2$
$V_1$- initial volume = $2.4\times {10}^3{m}^3$
$T_1$- initial temperature = $300K$

When the heat is supplied by the heater, the piston is compressed by 0.1 m. The reaction force of compression of spring is equal to $kx$ which acts on the piston or on the gas as
$F=kx=8000\times 0.1=800N$

Pressure exerted on the piston by the spring is
$\Delta P=\left(F\right)/\left(A\right)=\left(800\right)/(8\times {10}^{-3}=1\times {10}^{5}N//m^2$
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.,
$\Delta V=A\times x=(8\times {10}^{-3}\times (0.1)$
$=8\times {10}^{-4}{m}^3$
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
$\left(P_1{V}_1/T_1\right)=\left(P_2{V}_2/T_2\right)⟹T_2=\left(P_2{V}_2/P_1{V}_1\right)T_1$
$=300\times (2\times {10}^5\times 3.2\times {10}^{-3})/({10}^5\times 2.4\times {10}^{-3})$
$=800K$