Question

Two liters of an ideal gas at a pressure of \[10atm\] expands isothermally in a vacuum until its total volume is 10 liters. How much heat is absorbed in the expansion?
A.\[10J\]
B.\[8J\]
C.\[ - 18J\]
D.\[0J\]

Answer 1

Hint: According to the first law of thermodynamics, the total energy of the system is always constant, and heat absorbed or released will be the sum of internal energy of the system and work done.
Isothermal expansion: It is a thermodynamic process in which the temperature of the system always remains constant although the system is undergoing some changes.
Example: Melting of ice, Boiling of water.
Formula Used:
Work done by an ideal gas \[ \Rightarrow W = - {P_{ext}}\Delta V\]
The equation for the first law of thermodynamics \[ \Rightarrow \Delta Q = \Delta U + \Delta W{\text{ }}\]
\[ \Rightarrow Q = U + W\] (simplified form)

Complete answer:
Given that,
Initial volume, \[{V_1} = 2L\]
Final volume, \[{V_2} = 10L\]
Change in volume, \[\Delta V = {V_2} - {V_1} = 10-2 = 8{\text{ }}L\]
Internal pressure, \[{P_{in}} = 10atm\]
External pressure in a vacuum, \[{P_{ext}} = 0\]
Substituting the given data in the ideal gas equation.
\[W = - {P_{ext}}\Delta V = \left( { - 0{\text{ }} \times {\text{ }}8} \right) = 0J\]
Now according to the equation of the first law of thermodynamics \[ \Rightarrow Q = U + W\],
where \[U\] is internal energy and \[Q\] is the heat absorbed or released by the system and \[W\] is work done by or on the system.
In our case net work done is \[0J\] and internal energy is also \[0J\] since internal energy is directly proportional to temperature \[(\Delta U \propto \Delta T)\] .
Hence, we can say that heat absorbed in the expansion is
\[Q = 0J\]

Option D is correct among all.

Note:
For an ideal gas, work done is directly proportional to external pressure and change is volume. Whereas if external pressure is zero there is no external work. And, the system will undergo free expansion.