Question

If a gas is adiabatically compressed from state - I to state - II, suddenly by a single step, as shown in the figure then which of the following is true?

• A. $\Delta$H = 2$\gamma$ P$_0$V$_0$/$\gamma$-1
• B. $\Delta$H = -P$_0$V$_0$
• C. $\Delta$U = 3P$_0$V$_0$/$\gamma$-1
• D. $\Delta$U= -2 P$_0$V$_0$

Answer 1

Answer: (B)

$\Delta$H = -P$_0$V$_0$

Answer 2

The problem describes an adiabatic compression from state I to state II, suddenly by a single step. From the figure, state 1 has coordinates $(V_0, 2P_0)$ and state 2 has coordinates $(4V_0, P_0)$. Since the process is a compression, the initial state must have a larger volume than the final state, and the final state must have a higher pressure than the initial state. Looking at the figure, state 2 has a larger volume ($4V_0$) and lower pressure ($P_0$) compared to state 1 ($V_0$, $2P_0$). Therefore, for a compression, the initial state must be state 2 and the final state must be state 1. So, state I is state 2 and state II is state 1.

Initial state (state I): $V_i = V_2 = 4V_0$, $P_i = P_2 = P_0$. Final state (state II): $V_f = V_1 = V_0$, $P_f = P_1 = 2P_0$.

The process is a sudden single-step adiabatic compression. For a sudden single-step process, the work done by the gas is given by $W = P_{ext}(V_f - V_i)$, where $P_{ext}$ is the constant external pressure against which the process occurs. In a sudden compression, the external pressure is equal to the final pressure of the gas. So, $P_{ext} = P_f = P_1 = 2P_0$. The work done by the gas is $W = 2P_0 (V_1 - V_2) = 2P_0 (V_0 - 4V_0) = 2P_0 (-3V_0) = -6P_0V_0$. Since the process is adiabatic, $Q = 0$. According to the first law of thermodynamics, the change in internal energy is $\Delta U = Q - W = 0 - (-6P_0V_0) = 6P_0V_0$.

Now let's calculate the change in enthalpy, $\Delta H$. For an ideal gas, $\Delta H = \Delta U + \Delta (PV)$. $\Delta (PV) = P_f V_f - P_i V_i = P_1 V_1 - P_2 V_2 = (2P_0)(V_0) - (P_0)(4V_0) = 2P_0V_0 - 4P_0V_0 = -2P_0V_0$. $\Delta H = \Delta U + \Delta (PV) = 6P_0V_0 + (-2P_0V_0) = 4P_0V_0$.

Our calculated values are $\Delta U = 6P_0V_0$ and $\Delta H = 4P_0V_0$. None of the options match these values.

Let's re-examine the options and the problem. It is possible that the options are derived under some specific assumptions or there is a typo in the question or options.

Let's consider the possibility that the process is from state 1 to state 2 as shown in the figure, even though the text says "compressed". If the process is from state 1 to state 2, it is an expansion. Initial state: 1 $(V_1 = V_0, P_1 = 2P_0)$. Final state: 2 $(V_2 = 4V_0, P_2 = P_0)$. Sudden single-step adiabatic expansion. The external pressure is constant and equal to the final pressure of the gas. $P_{ext} = P_f = P_2 = P_0$. Work done by the gas $W = P_{ext}(V_f - V_i) = P_0 (V_2 - V_1) = P_0 (4V_0 - V_0) = P_0 (3V_0) = 3P_0V_0$. Since adiabatic, $Q = 0$. $\Delta U = Q - W = 0 - 3P_0V_0 = -3P_0V_0$. Change in enthalpy $\Delta H = \Delta U + \Delta (PV)$. $\Delta (PV) = P_f V_f - P_i V_i = P_2 V_2 - P_1 V_1 = (P_0)(4V_0) - (2P_0)(V_0) = 4P_0V_0 - 2P_0V_0 = 2P_0V_0$. $\Delta H = \Delta U + \Delta (PV) = -3P_0V_0 + 2P_0V_0 = -P_0V_0$.

Let's check the options again with $\Delta U = -3P_0V_0$ and $\Delta H = -P_0V_0$. Option 1: $\Delta H = 2\gamma P_0V_0/(\gamma-1)$. Option 2: $\Delta H = -P_0V_0$. This matches our calculated value of $\Delta H$ for an expansion from state 1 to state 2. Option 3: $\Delta U = 3P_0V_0/(\gamma-1)$. Option 4: $\Delta U = -2 P_0V_0$.

Since option 2 matches our calculation for an adiabatic expansion from state 1 to state 2, it is highly probable that the question meant adiabatic expansion from state I (which is state 1) to state II (which is state 2), despite saying "compressed".

Let's assume the question meant adiabatic expansion from state 1 to state 2. Initial state: 1 $(V_1 = V_0, P_1 = 2P_0)$. Final state: 2 $(V_2 = 4V_0, P_2 = P_0)$. Sudden single-step adiabatic expansion. External pressure is constant and equal to the final pressure, $P_{ext} = P_2 = P_0$. Work done by the gas $W = P_{ext}(V_2 - V_1) = P_0(4V_0 - V_0) = 3P_0V_0$. $\Delta U = Q - W = 0 - 3P_0V_0 = -3P_0V_0$. $\Delta H = \Delta U + \Delta (PV) = -3P_0V_0 + (P_2V_2 - P_1V_1) = -3P_0V_0 + (P_0(4V_0) - 2P_0V_0) = -3P_0V_0 + 4P_0V_0 - 2P_0V_0 = -P_0V_0$.

So, if the process is an adiabatic expansion from state 1 to state 2, then $\Delta H = -P_0V_0$. This matches option 2.

Let's verify if any other option can be correct under some interpretation. Consider option 3: $\Delta U = 3P_0V_0/(\gamma-1)$. If this were true, then $\Delta U$ is positive, which means internal energy increases. This happens during adiabatic compression. So, if the process is adiabatic compression, $\Delta U$ should be positive. If the process is adiabatic compression from state 2 to state 1, we calculated $\Delta U = 6P_0V_0$. If the process were reversible adiabatic compression from state 2 to state 1, then $P_2 V_2^\gamma = P_1 V_1^\gamma$. $P_0 (4V_0)^\gamma = 2P_0 V_0^\gamma$, so $4^\gamma = 2$, which is not possible for $\gamma > 1$. So, it's not a reversible adiabatic process.

Let's assume the question is correct in stating "adiabatically compressed from state - I to state - II" and the figure shows state I as 2 and state II as 1. Initial state: 2 $(4V_0, P_0)$. Final state: 1 $(V_0, 2P_0)$. Sudden single-step adiabatic compression. External pressure $P_{ext} = P_1 = 2P_0$. $W = -6P_0V_0$. $\Delta U = 6P_0V_0$. $\Delta H = 4P_0V_0$. None of the options match.

Given that option 2 matches the calculation for adiabatic expansion from state 1 to state 2, it is most likely that the question has a typo and it should be "adiabatically expanded from state - I to state - II". Assuming this interpretation, state I is state 1 and state II is state 2.

The final answer is $\Delta H = -P_0V_0$.