Three moles of an ideal gas are compressed isothermally from 60 L to 20 L using constant pressure of 5 atm. Heat exchange Q for the compression is – ____ Lit. atm.

The correct answer is 200.

To find the heat exchange, we can analyze the system using the principles of thermodynamics:

1. Identify the given values:
- Initial Volume ($V_1$) = 60 L
- Final Volume ($V_2$) = 20 L
- Constant External Pressure ($P_{\text{ext}}$) = 5 atm

2. Apply the First Law of Thermodynamics:
For an ideal gas undergoing an isothermal process, the temperature remains constant, which means there is no change in internal energy ($\Delta U$):
$\Delta U=0$
According to the first law of thermodynamics:
$\Delta U=Q+W$
where $Q$ is the heat exchange and $W$ is the work done on the system. Since $\Delta U=0$, we have:
$Q=-W$

3. Calculate the work done (W):
The work done during compression against a constant external pressure is given by:
$W=-{\mathrm{P<mtext>ext</mtext>}}_{}(V_2-V_1)$
Substituting the given values:
$W=-5\text{atm}\times (20\text{L}-60\text{L})$
$W=-5\text{atm}\times (-40\text{L})$
$W=200\text{L}·\text{atm}$

4. Calculate the heat exchange (Q):
Using the relationship $Q=-W$:
$Q=-200\text{L}·\text{atm}$

Thus, the heat exchange $Q$ for the compression is – 200 Lit. atm.