
A gas absorbs 200 J of heat and expands against the external pressure of 1.5 atm from a volume of 0.5 litres to 1.0 litre. Calculate the change in internal energy.
A. 124 J
B. 224 J
C. 114 J
D. 154 J
Answer
163.8k+ views
Hint: In a closed system, the Internal energy (U) is defined as the total energy of that system. In other words, the internal energy is the sum of the potential energy of the system and the kinetic energy of the system. The change in internal energy is represented by (\[\Delta U\]). By using the relation between internal energy, heat and work done we can determine the value of internal energy.
Formula used Work done is given as:
\[W = {P_{ext}}({V_2} - {V_1})\]
Where \[{P_{ext}}\]is the external pressure
\[{V_1}\]is initial volume
\[{V_2}\]is final volume
In a closed system, the internal energy (U) is given as:
\[\Delta U = Q - W\]
Where U is the change in internal energy of a system during a process
Q is the heat supplied
W is the mechanical work done
Complete step-by-step answer: Given Heat supplied Q=200 J
External pressure, \[{P_{ext}}\]=1.5 atm
Initial volume, \[{V_1} = 0.5L\]
Final volume, \[{V_1} = 1.0L\]
Work done, \[W = {P_{ext}}({V_2} - {V_1})\]
\[ = 1.5(1.0 - 0.5)\]
\[ = 0.75Latm\]
\[ = 0.75 \times 101.33Joule\]
\[W = 75.99Joule\]
As we know that \[\Delta U = Q - W\]
\[\Delta U = 200 J - 75.99 J = 124.0 J\]
Therefore, the change in internal energy is 124 J
Hence option A is the correct answer
Note: The internal energy represented as U is defined as the heat of the system. When the surrounding heat increases then the heat of the system decreases due to heat not being created nor destroyed. Hence heat is taken away from the system and makes it negative.
Formula used Work done is given as:
\[W = {P_{ext}}({V_2} - {V_1})\]
Where \[{P_{ext}}\]is the external pressure
\[{V_1}\]is initial volume
\[{V_2}\]is final volume
In a closed system, the internal energy (U) is given as:
\[\Delta U = Q - W\]
Where U is the change in internal energy of a system during a process
Q is the heat supplied
W is the mechanical work done
Complete step-by-step answer: Given Heat supplied Q=200 J
External pressure, \[{P_{ext}}\]=1.5 atm
Initial volume, \[{V_1} = 0.5L\]
Final volume, \[{V_1} = 1.0L\]
Work done, \[W = {P_{ext}}({V_2} - {V_1})\]
\[ = 1.5(1.0 - 0.5)\]
\[ = 0.75Latm\]
\[ = 0.75 \times 101.33Joule\]
\[W = 75.99Joule\]
As we know that \[\Delta U = Q - W\]
\[\Delta U = 200 J - 75.99 J = 124.0 J\]
Therefore, the change in internal energy is 124 J
Hence option A is the correct answer
Note: The internal energy represented as U is defined as the heat of the system. When the surrounding heat increases then the heat of the system decreases due to heat not being created nor destroyed. Hence heat is taken away from the system and makes it negative.
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