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 $300K$ and occupies a volume of $2.4\times {10}^{-3}{m}^3$ and the spring is in its relaxed (unstretched,uncompressed) state. The gas is heated by a small electric heater until the piston moves out slowly by $0.1m$. Calculate the final temperature of the gas and the heat supplied (in joules) by the heater. The force constant of the spring is $8000N/m$, and the atmospheric pressure $1.0\times 10$. 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 the lead wires of the heater. The heat capacity of the heater coil is negligible. Assume the spring to the massless.

Answer 1

$A=8\times {10}^{–3}{m}^2$

$\Delta x=0.1$

$T_1=300K$

$V_1=2.4\times {10}^{–3}{m}^3$

$V_2=V_1+A\Delta x=2.4\times {10}^{-3}+8\times {10}^{-3}\times 0.1=3.2\times {10}^{-3}{m}^3$

$K=8000N/m$

$T_2=?$

$P_1={10}^{5}N/m^2$

$P_2=P_1+\frac{k\Delta x}{A}={10}^5+\frac{8000\times 0.1}{8\times {10}^{-3}}=2\times {10}^{5}N/m^2$

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}⟹T_2=\frac{P_2{V}_2{T}_1}{P_1{V}_1}=\frac{2\times {10}^5\times 3.2\times {10}^{-3}\times 300}{{10}^5\times 2.4\times {10}^{-3}}=\frac{1920\times {10}^2}{2.4\times {10}^2}=800K$

Is the final temperature of the gas.

We know,

$\Delta U=\Delta Q-W⟹\Delta Q=\Delta U+W=\frac{3}{2}(P_2{V}_2-P_1{V}_1)+(P_{1}Ax+\frac{1}{2}K{x}^2)$

$\Delta Q=\frac{3}{2}\left[\right(2\times {10}^5\times 3.2\times {10}^{-3})-({10}^5\times 2.4\times {10}^{-3}\left)\right]+\left[\right({10}^5\times 8\times {10}^{-3}\times 0.1)+\frac{1}{2}(8000\times \left(0.1{)}^2\right)$

$\Delta Q=\frac{3}{2}(6.4\times {10}^2-2.4\times {10}^2)+(0.8\times {10}^2+40)$

$\Delta Q=\frac{3}{2}(4\times {10}^2)+(1.2\times {10}^2)$

$\Delta Q=7.2\times {10}^{2}J$