Question

Consider a piston–cylinder assembly as shown. The cylinder contains 1 mole of an ideal gas at 300 K, initially held at position $z_1$. After the stopper is removed, the piston rises suddenly against atmospheric pressure ($1.013\times10^{5}$ Pa) to a new position $z_2$. The cylinder walls are insulated. The heat capacity at constant volume $(C_V)$ is 12.5 J mol\textsuperscript{-1 K\textsuperscript{-1}. The cross-sectional area of the cylinder is $10^{-3}$ m\textsuperscript{2}. Assume the piston is weightless and frictionless. If $z_2 - z_1 = 1$ m, the final temperature of the gas (rounded to nearest integer) is _________ K.}

Hint:

For sudden (irreversible) adiabatic expansion, use $W = P_{\text{ext}}\Delta V$ and apply $\Delta U = -W$. Do NOT use reversible adiabatic relations.

Answer 1

Correct Answer: 291

Since the cylinder is insulated, the process is adiabatic.
But the piston moves suddenly, so the work done is against a constant external pressure:
\[ W = P_{\text{ext}}\, \Delta V \]
Given:
\[ P_{\text{ext}} = 1.013\times10^{5}\, \text{Pa} \]
\[ \Delta V = A (z_2 - z_1) = (10^{-3})(1) = 10^{-3}\, \text{m}^3 \]
Therefore:
\[ W = 1.013\times10^{5} \times 10^{-3} = 101.3\; \text{J} \]
[6pt] First law for adiabatic insulated system with work done by gas:
\[ \Delta U = -W \]
Internal energy change for an ideal gas:
\[ \Delta U = n C_V (T_2 - T_1) \]
Given: \(n = 1\), \(C_V = 12.5\) J mol\textsuperscript{-1}K\textsuperscript{-1}, \(T_1 = 300\) K.
Thus:
\[ 1(12.5)(T_2 - 300) = -101.3 \]
\[ T_2 - 300 = -8.104 \]
\[ T_2 = 291.896\; \text{K} \]
Rounded to nearest integer:
\[ \boxed{292\; \text{K}} \]