Thermodynamics Part 5

Concept:The first law of thermodynamics and work done by a gas at constant pressure are used to find the piston displacement.Explanation:Number of moles $n = 1$.Temperature rise $\Delta T = 4^\circ \mathrm{C} = 4 \ \mathrm{K}$.Heat supplied $\Delta Q = 126 \ \mathrm{J}$.For a monoatomic gas, change in internal energy $\Delta U = 3 n R \Delta T$.Substitute values: $\Delta U = 3 \times 1 \times 8.314 \times 4 = 99.768 \ \mathrm{J}$.From first law: $\Delta Q = \Delta U + \Delta W$, so $\Delta W = \Delta Q - \Delta U = 126 - 99.768 = 26.232 \ \mathrm{J}$.Piston area $A = 17 \ \mathrm{cm}^2 = 17 \times 10^{-4} \ \mathrm{m}^2$.Atmospheric pressure $P = 10^5 \ \mathrm{Pa}$, constant.Work done at constant pressure: $\Delta W = P \Delta V = P \cdot A \cdot x$, where $x$ is piston displacement in meters.Thus, $26.232 = 10^5 \times (17 \times 10^{-4}) \times x = 170 \times x$.Solve: $x = \frac{26.232}{170} \approx 0.1543 \ \mathrm{m}$.Convert to cm: $x = 0.1543 \times 100 = 15.43 \ \mathrm{cm} \approx 15.5 \ \mathrm{cm}$.Answer:The piston moves $15.5 \ \mathrm{cm}$, which corresponds to option C.