
A cylinder fitted with a piston contain 0.2 moles of air at temperature \[{27^ \circ }C.\]The piston is pushed so slowly that the air within the cylinder remains in thermal equilibrium with the surroundings. Find the approximate work done by the final volume is twice the initial volume
A. 549 J
B. 345 J
C. 453 J
D. 600 J
Answer
162.6k+ views
Hint:
No change in temperature hence it is an isothermal process. An isothermal process is a process where the temperature of the system remains constant. You can calculate work done of an isothermal process by putting the value P means pressure from the ideal gas law equation in the equation $W = pdV$and by solving it.
Complete step by step solution:
We know that ${0^0}C = 273K$. Hence ${27^0}C = 300K$.
By looking at the question carefully, we see that water is at \[300K\] and also the ice is at \[300K\]temperature. Hence, no change in temperature. So, the process is isothermal because only change in states happens. Let’s discuss work done in an isothermal process.
Boyle's law states that the pressure and volume of a gas are inversely proportional at a constant temperature.
Now let’s solve the question.
Work done in isothermal process is given by \[{W_{iso}} = 2.306nRTlo{g_{10}}\dfrac{{{V_2}}}{{{V_1}}}\] here ${V_1} = {\text{initial volume , }}{{\text{V}}_2} = {\text{final volume, R = universal gas constant , T = temperature,}}$
$n = {\text{number of moles}}$.
Here we are taking ${V_1} = V{\text{ and }}{{\text{V}}_2} = 2V$.
So now if we put the values as given in the question then we get
\[{W_{iso}} = 2.306nRTlo{g_{10}}\dfrac{{{V_2}}}{{{V_1}}}\]$ = 2.306 \times 8.31 \times 0.2 \times 300 \times {\log _{10}}\left( {\dfrac{{2V}}{{1V}}} \right){\text{joule}}$$ = 345Joule(approx)$
Hence the correct option is B.
Therefore, option (B) is the correct option.
Note:
Isothermal processes happen under a constant temperature. Work done of an isothermal process is \[{W_{iso}} = 2.306nRTlo{g_{10}}\dfrac{{{V_2}}}{{{V_1}}}\] . Alternatively, we can also write the work done in isothermal process as \[{W_{iso}} = 2.306nRTlo{g_{10}}\dfrac{{{P_1}}}{{{P_2}}}\]. If work is done by the system, then its value will be positive and if work is done on the system then its value will be negative.
No change in temperature hence it is an isothermal process. An isothermal process is a process where the temperature of the system remains constant. You can calculate work done of an isothermal process by putting the value P means pressure from the ideal gas law equation in the equation $W = pdV$and by solving it.
Complete step by step solution:
We know that ${0^0}C = 273K$. Hence ${27^0}C = 300K$.
By looking at the question carefully, we see that water is at \[300K\] and also the ice is at \[300K\]temperature. Hence, no change in temperature. So, the process is isothermal because only change in states happens. Let’s discuss work done in an isothermal process.
Boyle's law states that the pressure and volume of a gas are inversely proportional at a constant temperature.
Now let’s solve the question.
Work done in isothermal process is given by \[{W_{iso}} = 2.306nRTlo{g_{10}}\dfrac{{{V_2}}}{{{V_1}}}\] here ${V_1} = {\text{initial volume , }}{{\text{V}}_2} = {\text{final volume, R = universal gas constant , T = temperature,}}$
$n = {\text{number of moles}}$.
Here we are taking ${V_1} = V{\text{ and }}{{\text{V}}_2} = 2V$.
So now if we put the values as given in the question then we get
\[{W_{iso}} = 2.306nRTlo{g_{10}}\dfrac{{{V_2}}}{{{V_1}}}\]$ = 2.306 \times 8.31 \times 0.2 \times 300 \times {\log _{10}}\left( {\dfrac{{2V}}{{1V}}} \right){\text{joule}}$$ = 345Joule(approx)$
Hence the correct option is B.
Therefore, option (B) is the correct option.
Note:
Isothermal processes happen under a constant temperature. Work done of an isothermal process is \[{W_{iso}} = 2.306nRTlo{g_{10}}\dfrac{{{V_2}}}{{{V_1}}}\] . Alternatively, we can also write the work done in isothermal process as \[{W_{iso}} = 2.306nRTlo{g_{10}}\dfrac{{{P_1}}}{{{P_2}}}\]. If work is done by the system, then its value will be positive and if work is done on the system then its value will be negative.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Charging and Discharging of Capacitor

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
