Answer
397.2k+ views
Hint:To answer this question, you must recall the first law of thermodynamics. The first law of thermodynamics states that energy in the universe remains constant always, i.e., it is neither created nor destroyed.
Formula used:
First Law of Thermodynamics: $\Delta {\text{U}} = {\text{Q}} - {\text{W}}$
Where, $\Delta {\text{U}}$ is the internal energy of the given closed system
${\text{Q}}$ is the heat released or absorbed during the reversible isothermal expansion of the given gas.
And, ${\text{W}}$ is the work done during the reversible isothermal expansion of the given gas.
The work done in an isothermal reversible expansion is given by,
${\text{W}} = - nRT\ln \left( {\dfrac{{{V_2}}}{{{V_1}}}} \right) = - nRT\ln \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right)$
Where, $n$ is the number of moles of the ideal gas present in the given system
${V_1}$ and ${V_2}$ are the initial and final volumes of the system during the process of expansion.
And, ${P_1}$ and ${P_2}$ are the initial and final pressures of the system during the process of the expansion
Complete step by step answer:
It is given that the process is isothermal, i.e., there is no change in the temperature of the system. Since the internal energy of a system depends on the temperature, hence, we can conclude that the change in internal energy of the system will be zero.
From the equation of the first law of thermodynamics, we get, $\Delta {\text{U}} = {\text{Q}} - {\text{W}}$
$ \Rightarrow {\text{Q}} = - {\text{W}}$
We know that the work done in an isothermal reversible expansion is given by
${\text{W}} = - nRT\ln \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right)$
Substituting the values in the above equation, we get,
${\text{W}} = - \left( 1 \right) \times 8.314 \times 273 \times \ln \left( {\dfrac{1}{{0.1}}} \right)$
$ \Rightarrow {\text{W}} = - 8.314 \times 273 \times \ln 10$
Solving this:
$ \Rightarrow {\text{W}} = - 5226.23{\text{ J}}$
$ {\text{Q}} = 5226.23{\text{ Joules}}$
Note:
The first law of thermodynamics is an adapted form of the law of conservation of energy which has been modified so as to be applied in thermodynamic processes. It distinguishes two types of transfer of energy, as the heat transferred and the thermodynamic work done, during a process and relates them to Internal energy of the system which is a function of a system's state.
Formula used:
First Law of Thermodynamics: $\Delta {\text{U}} = {\text{Q}} - {\text{W}}$
Where, $\Delta {\text{U}}$ is the internal energy of the given closed system
${\text{Q}}$ is the heat released or absorbed during the reversible isothermal expansion of the given gas.
And, ${\text{W}}$ is the work done during the reversible isothermal expansion of the given gas.
The work done in an isothermal reversible expansion is given by,
${\text{W}} = - nRT\ln \left( {\dfrac{{{V_2}}}{{{V_1}}}} \right) = - nRT\ln \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right)$
Where, $n$ is the number of moles of the ideal gas present in the given system
${V_1}$ and ${V_2}$ are the initial and final volumes of the system during the process of expansion.
And, ${P_1}$ and ${P_2}$ are the initial and final pressures of the system during the process of the expansion
Complete step by step answer:
It is given that the process is isothermal, i.e., there is no change in the temperature of the system. Since the internal energy of a system depends on the temperature, hence, we can conclude that the change in internal energy of the system will be zero.
From the equation of the first law of thermodynamics, we get, $\Delta {\text{U}} = {\text{Q}} - {\text{W}}$
$ \Rightarrow {\text{Q}} = - {\text{W}}$
We know that the work done in an isothermal reversible expansion is given by
${\text{W}} = - nRT\ln \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right)$
Substituting the values in the above equation, we get,
${\text{W}} = - \left( 1 \right) \times 8.314 \times 273 \times \ln \left( {\dfrac{1}{{0.1}}} \right)$
$ \Rightarrow {\text{W}} = - 8.314 \times 273 \times \ln 10$
Solving this:
$ \Rightarrow {\text{W}} = - 5226.23{\text{ J}}$
$ {\text{Q}} = 5226.23{\text{ Joules}}$
Note:
The first law of thermodynamics is an adapted form of the law of conservation of energy which has been modified so as to be applied in thermodynamic processes. It distinguishes two types of transfer of energy, as the heat transferred and the thermodynamic work done, during a process and relates them to Internal energy of the system which is a function of a system's state.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)