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The volume of a gas decreases from \[500\;{\text{cc}}\] to \[300\;{\text{cc}}\] when a gas is compressed by average pressure of \[0.6\;{\text{atm}}\]. During the process, \[10\;{\text{J}}\] of heat is also liberated. The change in the internal energy is:
(A). \[22.16\;{\text{J}}\].
(B). \[{\text{12}}{\text{.56}}\;{\text{J}}\].
(C). \[{\text{2}}{\text{.16}}\;{\text{J}}\].
(D). \[{\text{101}}{\text{.3}}\;{\text{J}}\].

Last updated date: 29th Feb 2024
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IVSAT 2024
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Hint:In this question, the concept of the first law of thermodynamics will be used. Calculate the change in the volume of the gas. Then calculate the work done by the molecules of the system. Use the first law of thermodynamics to calculate the internal energy.

Complete step by step solution:
-Internal energy is the energy which is related with the random motion or the disorder of the molecules. Internal energy of the system depends upon the kinetic energy and the potential energy of the molecules that is due to motion.
-The internal energy of the system can be calculated by the first law of thermodynamics. This law states that energy can neither be created nor be destroyed but can be converted from one form to another.
-We can calculate the internal energy by using the formula,
\[ \Rightarrow \Delta U = q + W......\left( 1 \right)\]
Where, \[\Delta U\] is the change in internal energy, \[q\] is heat and \[W\] is the work done.
We are given heat liberated as,
\[ \Rightarrow q = - 10{\text{J}}\]
The pressure is given as,
\[ \Rightarrow {P_{ext}} = 0.6\;atm\]
Now we will calculate the change in volume,
\[ \Rightarrow \Delta V = 300 - 500\]
After simplification we will get
\[ \Rightarrow \Delta V = - 200\,{\text{cc}}\]
Now we will convert the volume into liter as
\[ \Rightarrow \Delta V = \dfrac{{ - 200}}{{1000}}\]
\[ \Rightarrow \Delta V = - 0.2\;{\text{L}}\]
Now, we will calculate the work done as
\[W = - {P_{ext}}.\Delta V\]
Now we substitute the values as,
   \Rightarrow W = - 0.6 \times \left( { - 0.2} \right) \\
   \Rightarrow W = 0.12 \\
Now we convert the work done into joule,
\[ \Rightarrow W = - 12.16J\]
Now we substitute the values in equation (1),
\[ \Rightarrow \Delta U = - 10 + 12.16\]
After calculation we will get,
\[\therefore \Delta U = 2.16\;{\text{J}}\]
So, change in the internal energy is \[2.16\;{\text{J}}\].
Therefore, the option (C) is correct.

As we know that the heat transfer and the work-done are the path function, that is it depends on the path of the process instead of the initial and final state. So, we take positive sign if the heat is transferred from surrounding to the system and negative value if the work is done from surrounding to the system.
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