One mole of an ideal monoatomic gas at `27^(@)` C expands adiabatically against constant external pressure of 1 atm from volume of 10 `dm^(3)` to a volume of `20 dm^(3)`.

Options

A. `W=+1013` Joule
B. `DeltaU=-1013` J
C. `DeltaH=+1688` J
D. The final temperature achieved is 291.8 K

Explanation

To solve the problem step by step, we will analyze the adiabatic expansion of one mole of an ideal monoatomic gas.

### Step 1: Understand the Given Data
- **Initial Temperature (T1)**: 27°C = 300 K (convert to Kelvin by adding 273)
- **Initial Volume (V1)**: 10 dm³
- **Final Volume (V2)**: 20 dm³
- **External Pressure (P_ext)**: 1 atm
- **Number of Moles (n)**: 1 mole

### Step 2: Calculate the Work Done (W)
The work done by the gas during an adiabatic expansion against a constant external pressure can be calculated using the formula:
\[ W = -P_{\text{ext}} \Delta V \]
Where:
- \(\Delta V = V_2 - V_1 = 20 \, \text{dm}^3 - 10 \, \text{dm}^3 = 10 \, \text{dm}^3\)

Substituting the values:
\[ W = -1 \, \text{atm} \times 10 \, \text{dm}^3 \]

Now, convert the work done from atm·dm³ to Joules:
1 atm·dm³ = 101.3 J
\[ W = -1 \times 10 \times 101.3 \, \text{J} = -1013 \, \text{J} \]

### Step 3: Calculate Change in Internal Energy (ΔU)
For an adiabatic process, the change in internal energy is equal to the work done on the system:
\[ \Delta U = W \]
Thus,
\[ \Delta U = -1013 \, \text{J} \]

### Step 4: Calculate Change in Enthalpy (ΔH)
The relationship between change in enthalpy and change in internal energy for an ideal gas is given by:
\[ \Delta H = \Delta U + \Delta nRT \]
Since we have one mole of gas, \(\Delta n = 0\) (no change in moles), and thus:
\[ \Delta H = \Delta U \]

However, we can also express the relationship in terms of specific heats:
\[ \frac{\Delta H}{\Delta U} = \frac{C_p}{C_v} \]
For a monoatomic ideal gas:
- \(C_v = \frac{3}{2}R\)
- \(C_p = C_v + R = \frac{5}{2}R\)

Thus,
\[ \frac{\Delta H}{\Delta U} = \frac{C_p}{C_v} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3} \]

Now substituting for \(\Delta H\):
\[ \Delta H = \frac{5}{3} \Delta U = \frac{5}{3} \times (-1013) = -1688.33 \, \text{J} \]

### Step 5: Calculate Final Temperature (T2)
Using the formula for change in internal energy:
\[ \Delta U = nC_v(T_2 - T_1) \]
Substituting the known values:
\[ -1013 = 1 \times \frac{3}{2}R (T_2 - 300) \]
Where \(R = 8.314 \, \text{J/(mol K)}\):
\[ -1013 = \frac{3}{2} \times 8.314 (T_2 - 300) \]
\[ -1013 = 12.471 (T_2 - 300) \]
\[ T_2 - 300 = \frac{-1013}{12.471} \]
\[ T_2 - 300 = -81.24 \]
\[ T_2 = 300 - 81.24 = 218.76 \, \text{K} \]

### Final Answers
- Work Done (W): -1013 J
- Change in Internal Energy (ΔU): -1013 J
- Change in Enthalpy (ΔH): -1688.33 J
- Final Temperature (T2): 218.76 K