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).
A perfect gas (0.1 mol) having \( \bar{C}_V = 1.50 \) R (independent of temperature) undergoes the above transformation from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is ____ J (nearest integer) [Given : R = 0.082 L atm K\(^{-1}\)]
A sample of n-octane (1.14 g) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5 kJ K\(^{-1}\). As a result of combustion, the temperature of the calorimeter increased by 5 K. The magnitude of the heat of combustion at constant volume is ___
Match List-I with List-II: List-I