2 mol of a monoatomic ideal gas with initial volume of 5 L and pressure 10 bar undergoes an irreversible adiabatic expansion against a constant final pressure of 1 bar. The final volume (in L) is ________.
(Given: R = $8.314 \times 10^{-2}$ L bar $mol^{-1}$ $K^{-1}$)
(rounded off to one decimal place)
Thermodynamics: Adiabatic Expansion Volume Calculation
Problem Setup
- Gas: 2 mol of monoatomic ideal gas.
- Initial State: $V_1 = 5$ L, $P_1 = 10$ bar.
- Process: Irreversible adiabatic expansion ($q=0$).
- External Pressure: Constant $P_{ext} = 1$ bar.
- Goal: Find final volume $V_2$.
Applying First Law of Thermodynamics
For an adiabatic process, the First Law ($\Delta U = q + w$) becomes $\Delta U = w$ since $q = 0$.
Work done ($w$) against a constant external pressure is given by $w = -P_{ext} \Delta V$. Substituting the values:
$w = -(1 \text{ bar}) (V_2 - 5 \text{ L})$Internal Energy Change
For a monoatomic ideal gas, the change in internal energy ($\Delta U$) is $n C_v \Delta T$. Using $C_v = \frac{3}{2}R$ for a monoatomic gas:
$\Delta U = n \frac{3}{2}R (T_2 - T_1)$Relating Temperature and Volume/Pressure
Using the ideal gas law ($PV=nRT$, so $T = \frac{PV}{nR}$):
$T_1 = \frac{P_1 V_1}{nR}$ $T_2 = \frac{P_2 V_2}{nR}$(Note: The final pressure of the gas $P_2$ equals the external pressure $P_{ext}$ at equilibrium).
Deriving the Equation for Volume
Equating $\Delta U = w$ and substituting the expressions for $\Delta U$ and $w$:
$n \frac{3}{2}R (T_2 - T_1) = -P_{ext} (V_2 - V_1)$Substitute $T_1$ and $T_2$:
$n \frac{3}{2}R \left(\frac{P_2 V_2}{nR} - \frac{P_1 V_1}{nR}\right) = -P_{ext} (V_2 - V_1)$Simplify the equation:
$\frac{3}{2} (P_2 V_2 - P_1 V_1) = -P_{ext} (V_2 - V_1)$Calculation
Substitute the known values ($P_1=10$ bar, $V_1=5$ L, $P_{ext}=1$ bar, $P_2=1$ bar):
$\frac{3}{2} ((1 \text{ bar}) V_2 - (10 \text{ bar})(5 \text{ L})) = -(1 \text{ bar}) (V_2 - 5 \text{ L})$Simplify and solve for $V_2$:
$1.5 (V_2 - 50) = -(V_2 - 5)$ $1.5 V_2 - 75 = -V_2 + 5$ $1.5 V_2 + V_2 = 5 + 75$ $2.5 V_2 = 80$ $V_2 = \frac{80}{2.5}$ $V_2 = 32$ LConclusion
The calculated final volume is 32 L, which lies within the specified range of 31.8 to 32.2 L.
Important Questions from First law of thermodynamics
- An ideal gas undergoes a process where $50J$ of work is done on the gas, and $30J$ of heat is removed from the gas.
Which of the following statements is true for the gas? When a gas is compressed suddenly then its temperature
During throttling process:
- Which among the following is correct option for adiabatic reversible expansion of an ideal gas?
From the above Carnot cycle undergone by an ideal gas, identify the processes in which the change in internal energy is NON-ZERO.
