Given:
- Mass of argon gas (m) = 10g
- Initial volume (V1) = 10L
- Final volume (V2) = 5L
- Temperature (T) = 27°C = 300K
- Atomic weight of argon = 40

To calculate:
- Heat transferred (q)
- Enthalpy change (ΔH)

Explanation:
1. The given process is isothermal, which means the temperature remains constant throughout the process. Therefore, the change in internal energy (ΔU) is zero.
2. The First Law of Thermodynamics states that ΔU = q - w, where ΔU is the change in internal energy, q is the heat transferred, and w is the work done on or by the system.
3. Since ΔU = 0, the equation becomes 0 = q - w.
4. For an isothermal process, the work done can be calculated using the equation w = -nRT ln(V2/V1), where n is the number of moles, R is the gas constant, and ln is the natural logarithm.
5. To calculate the number of moles (n), we can use the equation n = m/M, where m is the mass of the gas and M is the molar mass.
6. In this case, the mass of argon gas is given as 10g, and the molar mass of argon is 40 g/mol. Therefore, the number of moles (n) is 10g/40 g/mol = 0.25 mol.
7. The gas constant (R) is 8.314 J/(mol·K).
8. Substituting the values into the equation for work done, we get w = -0.25 mol * 8.314 J/(mol·K) * 300K * ln(5/10) = -311.8 J.
9. Since ΔU = 0, q = w. Therefore, q = -311.8 J.

Calculation of ΔH:
1. The enthalpy change (ΔH) can be calculated using the equation ΔH = q + PΔV, where P is the pressure and ΔV is the change in volume.
2. Since the process is reversible, the pressure remains constant. Therefore, PΔV = nRT ln(V2/V1).
3. Substituting the values, we get PΔV = 0.25 mol * 8.314 J/(mol·K) * 300K * ln(5/10) = -311.8 J.
4. Adding this value to q, we get ΔH = q + PΔV = -311.8 J + (-311.8 J) = -623.6 J.

Conclusion:
- The heat transferred (q) during the isothermal compression of 10g of argon gas from 10L to 5L at a temperature of 27°C is -311.8 J.
- The enthalpy change (ΔH) for this process is -623.6 J.