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Question: The pressure p, volume V and temperature T for a certain gas are related by \(p= \dfrac {AT-B{T}^{2}...

The pressure p, volume V and temperature T for a certain gas are related by p=AT−BT2Vp= \dfrac {AT-B{T}^{2}}{V}, where A and B are constants. The work done by the gas when the temperature changes from T1{T}_{1} to T2{T}_{2} while the pressure remains constant, is given by:
A.A(T2−T1)−B(T22−T12)A. A\left( { T }_{ 2 }-{ T }_{ 1 } \right) -B\left( { T }_{ 2 }^{ 2 }-{ T }_{ 1 }^{ 2 } \right)
B.A(T2−T1)V2−V1−B(T22−T12)V2−V1B. \dfrac { A\left( { T }_{ 2 }-{ T }_{ 1 } \right) }{ { V }_{ 2 }-{ V }_{ 1 } } -\dfrac { B\left( { T }_{ 2 }^{ 2 }-{ T }_{ 1 }^{ 2 } \right) }{ { V }_{ 2 }-{ V }_{ 1 } }
C.A(T2−T1)−B2(T22−T12)C. A\left( { T }_{ 2 }-{ T }_{ 1 } \right) -\dfrac {B}{2}\left( { T }_{ 2 }^{ 2 }-{ T }_{ 1 }^{ 2 } \right)
D.A(T2−T12)V2−V1D. \dfrac {A \left({T}_{2}-{T}_{1}^{2} \right)}{{V}_{2}-{V}_{1}}

Explanation

Solution

This problem can be solved using the concept of thermodynamics. Equation for pressure is given. Rearrange the equation and obtain an expression for pressure and volume. Then, differentiate the obtained equation with respect to T. We know, work done by a gas at constant pressure is given by W=∫pdVW= \int { pdV }. Substitute the equation in this formula and then integrate it from T1{T}_{1} to T2{T}_{2}. Evaluate it and obtain an expression for work done by the gas.

Complete step-by-step solution:
The pressure is given as,
p=AT−BT2Vp= \dfrac {AT-B{T}^{2}}{V} …(1)
Where,
p is the pressure
A and B are constants
T is the temperature
V is the volume

Rearranging equation. (1) we get,
pV=AT−BT2pV= AT- B{T}^{2}

Differentiating above equation with respect to T keeping pressure constant we get,
pdVdT=A−2BTp\dfrac { dV }{ dT } =A-2BT
⇒pdV=AdT−2BTdT\Rightarrow pdV= AdT-2BTdT …(2)

We know, work done by a gas at constant pressure is given by,
W=∫pdVW= \int { pdV }

Substituting equation. (2) in above equation we get,
W=∫AdT−2BTdTW= \int { AdT-2BTdT}

Integrating above equation from T1{T}_{1} to T2{T}_{2} we get,
W=∫T1T2AdT−2BTdTW= \int _{ { T }_{ 1 } }^{ { T }_{ 2 } }{ AdT-2BTdT }
⇒W=A∫T1T2dT−2B∫T1T2TdT\Rightarrow W= A\int _{ { T }_{ 1 } }^{ { T }_{ 2 } }{ dT-2B\int _{ { T }_{ 1 } }^{ { T }_{ 2 } }{ TdT } }
⇒W=A∫T1T2T−2B∫T1T2T22\Rightarrow W= A\int _{ { T }_{ 1 } }^{ { T }_{ 2 } }{ T-2B\int _{ { T }_{ 1 } }^{ { T }_{ 2 } }{ \dfrac { { T }^{ 2 } }{ 2 } } }
⇒W=A∫T1T2T−B∫T1T2T2\Rightarrow W= A\int _{ { T }_{ 1 } }^{ { T }_{ 2 } }{ T-B\int _{ { T }_{ 1 } }^{ { T }_{ 2 } }{ { T }^{ 2 } } }
⇒W=A(T2−T1)−B(T22−T12)\Rightarrow W= A\left( { T }_{ 2 }-{ T }_{ 1 } \right) -B\left( { T }_{ 2 }^{ 2 }-{ T }_{ 1 }^{ 2 } \right)
Thus, the work done by the gas is A(T2−T1)−B(T22−T12) A\left( { T }_{ 2 }-{ T }_{ 1 } \right) -B\left( { T }_{ 2 }^{ 2 }-{ T }_{ 1 }^{ 2 } \right).

So, the correct answer is option A i.e. A(T2−T1)−B(T22−T12) A\left( { T }_{ 2 }-{ T }_{ 1 } \right) -B\left( { T }_{ 2 }^{ 2 }-{ T }_{ 1 }^{ 2 } \right).

Note:
To solve these types of questions, students must be clear with the concepts of thermodynamics. Here, in this question, the pressure was kept constant, thus we got an expression for work done by the gas. If instead of pressure, volume was kept constant then we could have directly written the work done by the gas as zero. So, students must be aware of these concepts.