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.
Correct Answer: 200
Approach Solution - 1
As isothermal \( U = 0 \) and the process is irreversible:
\[Q = -W = -\left[-P_{\text{ext}}(V_2 - V_1)\right]\]
\[Q = 5 \times (20 - 60) = -200 \, \text{atm-L}\]
Given:
\[P_{\text{ext}} = 5 \, \text{atm}, \quad V_1 = 60 \, \text{L}, \quad V_2 = 20 \, \text{L}\]
Substituting the values:
\[Q = 5 \times (20 - 60) = -200 \, \text{atm-L}\]
Thus, the heat exchange for the compression is \( 200 \, \text{Lit. atm} \).
Approach Solution -2
We are dealing with the isothermal compression of an ideal gas. The given parameters are:
Number of moles, n = 3
Initial volume, V₁ = 60 L
Final volume, V₂ = 20 L
External (constant) pressure, Pext = 5 atm
Process type: Isothermal (temperature constant).
Step 2: Work done during the process
For a process carried out against a constant external pressure, the work done on the gas is:
w = −Pext(V₂ − V₁)
Substitute the given values:
w = −5(20 − 60) = −5(−40) = 200 L·atm
Since the work is done on the gas, w = +200 L·atm (positive for work done on the system).
Step 3: Apply the first law of thermodynamics
ΔU = q + w
For an isothermal process in an ideal gas, the internal energy change is zero (ΔU = 0) because internal energy depends only on temperature, which is constant.
Hence,
0 = q + w ⇒ q = −w
Thus,
q = −200 L·atm
The negative sign indicates that heat is released (i.e., the system loses heat to the surroundings).
Step 4: Final result
The heat exchange during the isothermal compression is:
Q = −200 L·atm
Final Answer:
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