A real gas has critical temperature and critical pressure as `40^(@)C` and 10 atm respectively, then liquification of gas is possible at

Options

A. `50^(@)C` and 8 atm
B. `45^(@)C` and 8 atm
C. `25^(@)C` and 12 atm
D. `45^(@)C` and 12 atm

Answer

To determine the conditions under which liquefaction of a real gas is possible, we need to consider the critical temperature and critical pressure of the gas. The given values are:

- Critical Temperature (T_c) = 40°C
- Critical Pressure (P_c) = 10 atm

### Step-by-Step Solution:

1. **Understanding Critical Temperature and Pressure**:
- The critical temperature is the maximum temperature at which a substance can exist as a liquid. Above this temperature, the substance cannot be liquefied regardless of the pressure applied.
- The critical pressure is the minimum pressure required to liquefy a gas at its critical temperature.

2. **Condition for Liquefaction**:
- For a gas to be liquefied, the temperature must be below the critical temperature (T < T_c) and the pressure must be above the critical pressure (P > P_c).

3. **Applying the Conditions**:
- Since the critical temperature is 40°C, any temperature below this will allow for the possibility of liquefaction. Therefore, we need a temperature less than 40°C.
- The critical pressure is 10 atm, so we need a pressure greater than 10 atm for liquefaction to occur.

4. **Conclusion**:
- Liquefaction of the gas is possible at temperatures less than 40°C and pressures greater than 10 atm.

### Final Answer:
Liquefaction of the gas is possible at temperatures below 40°C and pressures above 10 atm.