
A diatomic gas \[\left( {\gamma = 1.4} \right)\] does \[{\text{200 J}}\] of work when it is expanded isobarically. Find the heat given to the gas in the process.
A. \[{\text{500 J}}\]
B. \[{\text{700 J}}\]
C. \[{\text{600 J}}\]
D. \[{\text{900 J}}\]
Answer
570k+ views
Hint: First of all, we need to use the equation of first law of thermodynamics. Then we will find an equation which relates specific heat in constant volume and ratio of specific heats. We will make necessary modifications followed by substitution of required values. We will manipulate accordingly and obtain the required result.
Complete step by step answer:
Here,
From the First law of thermodynamics, we know that,
Energy can be changed from one form to another, but it cannot be created or destroyed. The total amount of energy and matter in the Universe remains constant, merely changing from one form to another.
So,
\[Q = p\Delta V + n{C_{\text{v}}}\Delta T\] …… (1)
Where,
\[Q\] indicates the heat added.
\[p\] indicates the pressure.
\[\Delta V\] indicates the change in volume.
\[n\] indicates the number of moles.
\[{C_{\text{v}}}\] indicates specific heat in constant volume.
\[\Delta T\] indicates the change in temperature.
Here,
We know that,
\[{C_{\text{v}}} = \dfrac{R}{{\gamma - 1}}\]
Where,
\[R\] indicates the gas constant.
Therefore,
We can write the equation (1) as,
\[Q = p\Delta V + n\dfrac{R}{{\gamma - 1}}\Delta T\]
Where,
\[\gamma \] indicates ratio of specific heats.
Again, we can write:
\[p\Delta V = nR\Delta T\]
After simplification, we get,
$Q = p\Delta V + n\dfrac{R}{{\gamma - 1}}\Delta T \\
\implies Q = p\Delta V\left( {1 + \dfrac{1}{{\gamma - 1}}} \right) \\$
Given,
\[p\Delta V = 200\]
\[\gamma - 1 = 0.4\]
Therefore,
$Q = p\Delta V\left( {1 + \dfrac{1}{{\gamma - 1}}} \right) \\
\implies Q = 200\left( {1 + \dfrac{1}{{0.4}}} \right) \\
\implies Q = 200 \times 3.5 \\
\therefore Q = 700\,{\text{J}} \\$
Hence,
The heat given to the gas in the process is \[{\text{700 J}}\] .
So, the correct answer is “Option B”.
Note:
The first law of thermodynamics is a variant of the law of energy conservation, modified for thermodynamic processes, distinguishing between two forms of energy transfer, such as heat and thermodynamic work, and comparing them to a function called internal energy of the state of the body. It is possible to shift energy from one form to another, but it cannot be created or destroyed. In the Universe the total sum of energy and matter remains constant, varying merely from one form to another.
The second law of thermodynamics states that the entropy of any isolated system always increases. The third law of thermodynamics states that, as the temperature reaches absolute zero, the entropy of a system reaches a constant value.
When a reaction happens, energy is transferred from one object to another. Energy comes in several forms and can be transmitted as heat, light, or motion from one entity to another.
Complete step by step answer:
Here,
From the First law of thermodynamics, we know that,
Energy can be changed from one form to another, but it cannot be created or destroyed. The total amount of energy and matter in the Universe remains constant, merely changing from one form to another.
So,
\[Q = p\Delta V + n{C_{\text{v}}}\Delta T\] …… (1)
Where,
\[Q\] indicates the heat added.
\[p\] indicates the pressure.
\[\Delta V\] indicates the change in volume.
\[n\] indicates the number of moles.
\[{C_{\text{v}}}\] indicates specific heat in constant volume.
\[\Delta T\] indicates the change in temperature.
Here,
We know that,
\[{C_{\text{v}}} = \dfrac{R}{{\gamma - 1}}\]
Where,
\[R\] indicates the gas constant.
Therefore,
We can write the equation (1) as,
\[Q = p\Delta V + n\dfrac{R}{{\gamma - 1}}\Delta T\]
Where,
\[\gamma \] indicates ratio of specific heats.
Again, we can write:
\[p\Delta V = nR\Delta T\]
After simplification, we get,
$Q = p\Delta V + n\dfrac{R}{{\gamma - 1}}\Delta T \\
\implies Q = p\Delta V\left( {1 + \dfrac{1}{{\gamma - 1}}} \right) \\$
Given,
\[p\Delta V = 200\]
\[\gamma - 1 = 0.4\]
Therefore,
$Q = p\Delta V\left( {1 + \dfrac{1}{{\gamma - 1}}} \right) \\
\implies Q = 200\left( {1 + \dfrac{1}{{0.4}}} \right) \\
\implies Q = 200 \times 3.5 \\
\therefore Q = 700\,{\text{J}} \\$
Hence,
The heat given to the gas in the process is \[{\text{700 J}}\] .
So, the correct answer is “Option B”.
Note:
The first law of thermodynamics is a variant of the law of energy conservation, modified for thermodynamic processes, distinguishing between two forms of energy transfer, such as heat and thermodynamic work, and comparing them to a function called internal energy of the state of the body. It is possible to shift energy from one form to another, but it cannot be created or destroyed. In the Universe the total sum of energy and matter remains constant, varying merely from one form to another.
The second law of thermodynamics states that the entropy of any isolated system always increases. The third law of thermodynamics states that, as the temperature reaches absolute zero, the entropy of a system reaches a constant value.
When a reaction happens, energy is transferred from one object to another. Energy comes in several forms and can be transmitted as heat, light, or motion from one entity to another.
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