Question

An ideal monatomic 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 300 K and occupies a volume of $2.4\times {10}^{-3}{m}^3$ and the spring is in its relaxed state. The gas is heated by a small heater until the piston moves slowly by 0.1 m. Calculate the final temperature of the gas. The force constant of the spring is 8000 N/m, and the atmospheric pressure is $1.0\times {10}^{5}N/m^2$. 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. Neglect any heat - loss through the lead wires of the heater. The heat capacity of the heater coil is negligible.

• A. 370 K
• B. 700 K
• C. 298 K
• D. 800 K

Answer 1

The correct option is D 800 K

Initially the spring is in its relaxed state. So, the pressure of the gas equals the atmospheric pressure.
Initial pressure = $P_1=1.0\times {10}^{5}N/m^2$.
Final pressure = $P_2=P_1+\frac{kx}{A}$
$=1.0\times {10}^{5}N/m^2+\frac{\left(8000N/m\right)\left(0.1m\right)}{8.0\times {10}^{-3}{m}^2}=2.0\times {10}^{5}N/m^2.\text{Final volume}=V_2=V_1+Ax=2.4\times {10}^{-3}{m}^3+8.0\times {10}^{-3}{m}^2\times 0.1m=3.2\times {10}^{-3}{m}^3.Using\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2},T_2=\frac{P_2{V}_2}{P_1{V}_1}{T}_1=\frac{(2.0\times {10}^{5}N/m^2)(3.2\times {10}^{-3}{m}^3)}{(1.0\times {10}^{5}N/m^2)(2.4\times {10}^{-3}{m}^3)}\times 300K=800K$