At `27^(@)C`, one mole of an ideal gas is compressed isothermally and reversibly from a pressure of `2` atm to `10` atm. Calculate `DeltaU` and `q`.

Options

A. `0,-965.84cal`
B. `0,-965.84"cal",-865.58"cal"`
C. `+865.58"cal",-865.58"cal"`
D. `0,-865.58"cal"`

Answer

To solve the problem of calculating \(\Delta U\) and \(q\) for the isothermal compression of one mole of an ideal gas from a pressure of 2 atm to 10 atm at a temperature of 27°C, we can follow these steps:

### Step 1: Identify the Given Values
- Number of moles (\(n\)) = 1 mole
- Initial pressure (\(P_1\)) = 2 atm
- Final pressure (\(P_2\)) = 10 atm
- Temperature (\(T\)) = 27°C = 300 K (since \(T(K) = T(°C) + 273\))
- Gas constant (\(R\)) = 2 cal/(mol·K)

### Step 2: Calculate the Work Done (W)
For an isothermal process, the work done on the gas can be calculated using the formula:
\[ W = -nRT \ln\left(\frac{P_2}{P_1}\right) \]
Substituting the known values:
\[ W = -1 \times 2 \times 300 \times \ln\left(\frac{10}{2}\right) \]
Calculating \(\ln\left(\frac{10}{2}\right) = \ln(5)\):
\[ W = -600 \times \ln(5) \]
Using \(\ln(5) \approx 1.609\):
\[ W = -600 \times 1.609 \approx -965.4 \text{ cal} \]

### Step 3: Calculate \(\Delta U\)
For an ideal gas undergoing an isothermal process, the change in internal energy (\(\Delta U\)) is given by:
\[ \Delta U = nC_v\Delta T \]
Since the process is isothermal, \(\Delta T = 0\):
\[ \Delta U = 0 \]

### Step 4: Calculate Heat Transfer (q)
Using the first law of thermodynamics:
\[ \Delta U = q + W \]
Rearranging gives:
\[ q = \Delta U - W \]
Substituting the values:
\[ q = 0 - (-965.4) = 965.4 \text{ cal} \]

### Final Results
- \(\Delta U = 0\)
- \(q \approx 965.4 \text{ cal}\)

### Summary
The final answers are:
- \(\Delta U = 0\)
- \(q \approx 965.4 \text{ cal}\)

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