
The work is done when a gas is compressed by an average pressure of 0.50 atm to decrease its volume from $400c{m^3}$to $200c{m^3}$.
A. 10.13J
B. 20.13J
C. 30.13J
D. 40.13J
Answer
563.7k+ views
Hint: A type of mechanical work done is said to be done when a gas expands or compresses against the external pressure and it is also known as pressure-volume work. Based on whether the heat is absorbed or emitted we can say work done is negative and positive respectively.
Complete step by step answer:
Here given that average pressure by which the gas is compressed is 0.50atm and And the initial volume of the gas $(V_1)$ is 400$cm^3$ and the final volume of the gas $(V_2)$ is 200$cm^3$.The pressure-volume work done for compression of a gas is given as:
$W = - P\Delta V$ (1)
Where $\Delta V = {V_2} - {V_1}$ (2)
On putting the given values in equation (2), we get:
$\Delta V = 200 - 400 = - 200c{m^3}$
Now substituting the value of $\Delta V$in equation (1), the work by compression of gas will be:
$W = - 0.5 \times - 200$
$ \Rightarrow W = \dfrac{{ - 05 \times - 200}}{{10}} = 100atm{\text{ }}c{m^3}$
As we know $1000c{m^3} = 1Ltr$
Then $100{\text{ }}atm{\text{ }}c{m^3} = 100 \times {10^{ - 3}}atm{\text{ }}Ltr$
Also $1\,atm{\text{ }}Ltr = 101.325J$
Thus $100 \times {10^{ - 3}}atmLtr = 100 \times {10^{ - 3}} \times 101.325J$
So, the correct answer is Option A.
Additional Information:
The relationship between work done and enthalpy change is that at constant pressure, if the work done results in change in volume then the enthalpy of the system is equal to the heat transferred to the system. The factors on which the enthalpy of the system depends are:
Physical state of the reactants and products
Temperature and pressure of the reaction
Quantity of the reactants used.
Note: As per thermodynamics, work done can be defined as mechanical or non-mechanical work. The examples of mechanical work are work done in flowing, rotating the shaft or while displacing. Non-mechanical work is categorised as work done in current flow, chemical reaction, etc.
Complete step by step answer:
Here given that average pressure by which the gas is compressed is 0.50atm and And the initial volume of the gas $(V_1)$ is 400$cm^3$ and the final volume of the gas $(V_2)$ is 200$cm^3$.The pressure-volume work done for compression of a gas is given as:
$W = - P\Delta V$ (1)
Where $\Delta V = {V_2} - {V_1}$ (2)
On putting the given values in equation (2), we get:
$\Delta V = 200 - 400 = - 200c{m^3}$
Now substituting the value of $\Delta V$in equation (1), the work by compression of gas will be:
$W = - 0.5 \times - 200$
$ \Rightarrow W = \dfrac{{ - 05 \times - 200}}{{10}} = 100atm{\text{ }}c{m^3}$
As we know $1000c{m^3} = 1Ltr$
Then $100{\text{ }}atm{\text{ }}c{m^3} = 100 \times {10^{ - 3}}atm{\text{ }}Ltr$
Also $1\,atm{\text{ }}Ltr = 101.325J$
Thus $100 \times {10^{ - 3}}atmLtr = 100 \times {10^{ - 3}} \times 101.325J$
So, the correct answer is Option A.
Additional Information:
The relationship between work done and enthalpy change is that at constant pressure, if the work done results in change in volume then the enthalpy of the system is equal to the heat transferred to the system. The factors on which the enthalpy of the system depends are:
Physical state of the reactants and products
Temperature and pressure of the reaction
Quantity of the reactants used.
Note: As per thermodynamics, work done can be defined as mechanical or non-mechanical work. The examples of mechanical work are work done in flowing, rotating the shaft or while displacing. Non-mechanical work is categorised as work done in current flow, chemical reaction, etc.
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