Question

A real gas is subjected to an adiabatic process from (4bar, 40L) to (6bar, 25L) against a constant pressure in a single step. The enthalpy change for the process (in bar litre) is x. The value of $\frac{x}{10}$ is __

Answer 1

8

Answer 2

The process is an adiabatic process from state 1 to state 2 in a single step against a constant external pressure.

Initial state: $(P_1, V_1) = (4 \text{ bar}, 40 \text{ L})$ Final state: $(P_2, V_2) = (6 \text{ bar}, 25 \text{ L})$

The process is adiabatic, so $q = 0$.

The process is carried out in a single step against a constant external pressure, let's call it $P_{ext}$. The work done by the system is given by $w = -P_{ext} (V_2 - V_1)$. For a single-step irreversible compression from state 1 to state 2 with final pressure $P_2$, the constant external pressure is equal to the final pressure, i.e., $P_{ext} = P_2$. So, $P_{ext} = 6 \text{ bar}$.

The work done is $w = -P_{ext} (V_2 - V_1) = -6 \text{ bar} (25 \text{ L} - 40 \text{ L}) = -6 \text{ bar} (-15 \text{ L}) = 90 \text{ bar L}$.

According to the first law of thermodynamics, $\Delta U = q + w$. Since the process is adiabatic, $q = 0$. So, $\Delta U = w = 90 \text{ bar L}$.

The enthalpy change is defined as $\Delta H = \Delta U + \Delta (PV)$. $\Delta (PV) = P_2 V_2 - P_1 V_1$. $P_1 V_1 = (4 \text{ bar})(40 \text{ L}) = 160 \text{ bar L}$. $P_2 V_2 = (6 \text{ bar})(25 \text{ L}) = 150 \text{ bar L}$. $\Delta (PV) = 150 \text{ bar L} - 160 \text{ bar L} = -10 \text{ bar L}$.

The enthalpy change is $\Delta H = \Delta U + \Delta (PV) = 90 \text{ bar L} + (-10 \text{ bar L}) = 80 \text{ bar L}$. The enthalpy change for the process is given as $x$. So, $x = 80 \text{ bar litre}$.

We are asked to find the value of $\frac{x}{10}$. $\frac{x}{10} = \frac{80}{10} = 8$.