Question

An ideal monoatomic gas is confined in a cylinder by a spring-loaded piston of cross-sectional area $8\times {10}^{-3}{\text{m}}^2$. Initially the gas is at $300\text{K}$ and occupies a volume of $2.4\times {10}^{-3}{\text{m}}^3$. The spring is in a relaxed state. The gas is heated by a small heater coil $H$. The force constant of the spring is $8000\text{N/m}$ and the atmospheric pressure is $1.0\times {10}^5\text{Pa}$. The cylinder and piston are thermally insulated. The piston and the spring are massless and there is no friction between the piston and cylinder. There is no heat loss through the heater coil wire and thermal 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 $0.1\text{m}$, then the final temperature is -

• A. $400\text{K}$
• B. $800\text{K}$
• C. $1200\text{K}$
• D. $300\text{K}$

Answer 1

The correct option is B $800\text{K}$

Given:
Cross sectional area of piston, $A=8\times {10}^{-3}{\text{m}}^2$
Initial temperature, $T_1=300\text{K}$
Initial volume, $V_1=2.4\times {10}^{-3}{\text{m}}^3$
Force constant of spring, $k=8000\text{N/m}$
Atmospheric pressure, $P_o=1.0\times {10}^5\text{Pa}$
Distance moved out by piston, $x=0.1\text{m}$
To find:
Final temperature, $T_2=?$

Initial pressure will be equal to the atmospheric pressure as spring is in natural state.
$\therefore P_1=P_0=1.0\times {10}^5\text{Pa}$
And, final pressure will balance the pressure due to atmosphere and spring.
$\therefore P_2=\frac{kx}{A}+P_0=\frac{8000\times 0.1}{8\times {10}^{-3}}+1.0\times {10}^5$
$=2\times {10}^5\text{Pa}$
Also, final volume will be equal to initial volume plus increase in volume.
$\Rightarrow V_2=V_1+Ax=2.4\times {10}^{-3}+8\times {10}^{-3}\times 0.1$
$=3.2\times {10}^{-3}{\text{m}}^3$

We know that, ideal gas equation is given by
$PV=nRT$
As number of moles will remain constant, so,
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$\Rightarrow T_2=\frac{P_2{V}_2{T}_1}{P_1{V}_1}$
$\Rightarrow T_2=\frac{2\times {10}^5\times 3.2\times {10}^{-3}\times 300}{1.0\times {10}^5\times 2.4\times {10}^{-3}}$
$\Rightarrow T_2=800\text{K}$