The q value and work done in isothermal reversible expansion of one mole of an ideal gas from an initial pressure of 10 bar to final pressure of 1 bar at constant temperature 273K are:a.) 5.22 KJ, -5.22 KJb.) -5.22 KJ, 5.22 KJc.) 5.22 KJ, 5.22 KJd.) -5.22 KJ, -5.22 KJ

To solve this question, we need to calculate the heat \( q \) and the work done \( W \) for an isothermal reversible expansion of an ideal gas.

### Given:
- One mole of ideal gas (\( n = 1 \))
- Initial pressure (\( P_i = 10 \, \text{bar} \))
- Final pressure (\( P_f = 1 \, \text{bar} \))
- Temperature (\( T = 273 \, \text{K} \))

### Formula for Work Done in Isothermal Expansion:
For an isothermal reversible expansion, the work done by an ideal gas is given by:

\[
W = - nRT \ln \left( \frac{V_f}{V_i} \right)
\]

Using the ideal gas law, we know that \( PV = nRT \), so:

\[
\frac{V_f}{V_i} = \frac{P_i}{P_f}
\]

Thus, the equation for work becomes:

\[
W = - nRT \ln \left( \frac{P_i}{P_f} \right)
\]

Substitute the values:

- \( n = 1 \)
- \( R = 8.314 \, \text{J/mol·K} \)
- \( T = 273 \, \text{K} \)
- \( P_i = 10 \, \text{bar} \)
- \( P_f = 1 \, \text{bar} \)

\[
W = - (1)(8.314)(273) \ln \left( \frac{10}{1} \right)
\]

First, calculate the natural logarithm:

\[
\ln(10) \approx 2.303
\]

Now calculate the work done:

\[
W = - (8.314)(273)(2.303) = -5216.3 \, \text{J} = -5.22 \, \text{kJ}
\]

### Heat Exchange \( q \):
In an isothermal process for an ideal gas, the change in internal energy (\( \Delta U \)) is zero because internal energy depends only on temperature, and the temperature is constant. Therefore, from the first law of thermodynamics:

\[
\Delta U = 0 = q + W
\]

So, \( q = -W \).

Thus, the heat \( q \) is:

\[
q = 5.22 \, \text{kJ}
\]

### Conclusion:
- Heat \( q = 5.22 \, \text{kJ} \)
- Work \( W = -5.22 \, \text{kJ} \)

The correct answer is:
**a.) 5.22 KJ, -5.22 KJ**