Work done ($W$) in an isothermal reversible expansion of an ideal gas is given by:
\[W = -nRT \ln \left( \frac{V_2}{V_1} \right)\]
Given:
\[n = 1 \, \text{mol}, \quad T = 18^\circ \text{C} = 18 + 273.15 = 291.15 \, \text{K}\]
\[V_1 = 10 \, \text{L}, \quad V_2 = 100 \, \text{L}\]
Substitute the values:
\[W = -1 \times 0.08206 \times 291.15 \times \ln \left( \frac{100}{10} \right)\]
\[W \approx -1 \times 0.08206 \times 291.15 \times \ln(10)\]
\[W \approx -55.0128 \, \text{L atm}\]
The work done by the system is approximately $-55 \, \text{L atm}$ (rounded to nearest integer).
The left and right compartments of a thermally isolated container of length $L$ are separated by a thermally conducting, movable piston of area $A$. The left and right compartments are filled with $\frac{3}{2}$ and 1 moles of an ideal gas, respectively. In the left compartment the piston is attached by a spring with spring constant $k$ and natural length $\frac{2L}{5}$. In thermodynamic equilibrium, the piston is at a distance $\frac{L}{2}$ from the left and right edges of the container as shown in the figure. Under the above conditions, if the pressure in the right compartment is $P = \frac{kL}{A} \alpha$, then the value of $\alpha$ is ____
If \[ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6 \], then f(1) is equal to: