Answer
Verified
391.7k+ views
Hint: Consider an isolated non-conducting cylinder, with a non-conducting piston and then pull the piston by a little amount outward which will cause a change in the physical conditions of the gas, and then find the work done to make the change.
Complete step by step solution:
Suppose we have a one-gram molecule of a perfect gas which has been enclosed in a non-conducting cylinder having a non-conducting piston. The gas will expand, if we move the piston slowly outwards and hence will do some work without any energy being supplied from outside and thus adiabatic expansion will take place. And since the energy hasn’t been used from outside, the temperature of the gas in the cylinder will fall.
Let us say the initial temperature, pressure and volume of the cylinder be $T_1$, $P_1$ and $V_1$ and in its final state be $T_2$, $P_2$ and $V_2$.
Now, let the cross-sectional area of the piston be A and we move the piston through a small distance dx which makes the gas expand by a volume of dV. And for small expansion, the pressure change will be almost the same, say P.
Now, for an adiabatic change $PV^{\gamma}=K$ (constant}
Therefore, $P=\dfrac{K}{V^{\gamma}}$
Thus, the work done, $W= \int_{V_1}^{V_2}PdV$
$\implies W=\int_{V_1}^{V_2}\dfrac{K}{V^{\gamma}}dV=K\int_{V_1}^{V_2}V^{-\gamma}dV$
$\therefore, W=K\left|\dfrac{V^{1-\gamma}}{1-\gamma}\right|=\dfrac{K}{1-\gamma}({V_2}^{1-\gamma}-{V_1}^{1-\gamma})$
Since, in adiabatic expansion, we know that
${P_1}{V_1}^{\gamma}={P_2}{V_2}^{\gamma}=k$
Thus, $W=\left(\dfrac{1}{1-\gamma}\right)({P_2}{V_2}^{\gamma}{V_2}^{1-\gamma}-{P_1}{V_1}^{\gamma}{V_1}^{1-\gamma})$
$\implies W=\left[\dfrac{1}{1-\gamma}\right][{P_2}{V_2}-{P_1}{V_1}]$
Using, $PV=RT$, we can also write the work done as,
$W=\left[\dfrac{R}{1-\gamma}\right][T_2 - T_1]$
This is the equation for work done in adiabatic expansion.
Note: We can also derive the expression for the work done by graphical method using a PV curve for the change in pressure and volume during an adiabatic expansion, then the area under the curve will give us the required work done.
Complete step by step solution:
Suppose we have a one-gram molecule of a perfect gas which has been enclosed in a non-conducting cylinder having a non-conducting piston. The gas will expand, if we move the piston slowly outwards and hence will do some work without any energy being supplied from outside and thus adiabatic expansion will take place. And since the energy hasn’t been used from outside, the temperature of the gas in the cylinder will fall.
Let us say the initial temperature, pressure and volume of the cylinder be $T_1$, $P_1$ and $V_1$ and in its final state be $T_2$, $P_2$ and $V_2$.
Now, let the cross-sectional area of the piston be A and we move the piston through a small distance dx which makes the gas expand by a volume of dV. And for small expansion, the pressure change will be almost the same, say P.
Now, for an adiabatic change $PV^{\gamma}=K$ (constant}
Therefore, $P=\dfrac{K}{V^{\gamma}}$
Thus, the work done, $W= \int_{V_1}^{V_2}PdV$
$\implies W=\int_{V_1}^{V_2}\dfrac{K}{V^{\gamma}}dV=K\int_{V_1}^{V_2}V^{-\gamma}dV$
$\therefore, W=K\left|\dfrac{V^{1-\gamma}}{1-\gamma}\right|=\dfrac{K}{1-\gamma}({V_2}^{1-\gamma}-{V_1}^{1-\gamma})$
Since, in adiabatic expansion, we know that
${P_1}{V_1}^{\gamma}={P_2}{V_2}^{\gamma}=k$
Thus, $W=\left(\dfrac{1}{1-\gamma}\right)({P_2}{V_2}^{\gamma}{V_2}^{1-\gamma}-{P_1}{V_1}^{\gamma}{V_1}^{1-\gamma})$
$\implies W=\left[\dfrac{1}{1-\gamma}\right][{P_2}{V_2}-{P_1}{V_1}]$
Using, $PV=RT$, we can also write the work done as,
$W=\left[\dfrac{R}{1-\gamma}\right][T_2 - T_1]$
This is the equation for work done in adiabatic expansion.
Note: We can also derive the expression for the work done by graphical method using a PV curve for the change in pressure and volume during an adiabatic expansion, then the area under the curve will give us the required work done.
Recently Updated Pages
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
A rainbow has circular shape because A The earth is class 11 physics CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
The Constitution of India was adopted on A 26 November class 10 social science CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE