One mole of a vander Waalsâ gas obeying the equation (P+(a/V2))(V-b)=RT undergoes the quasi-static cyclic process which is shown in the P-V diagram. The net heat absorbed by the gas in this process is

Question

One mole of a vander Waalsâ gas obeying the equation (P+(a/V2))(V-b)=RT undergoes the quasi-static cyclic process which is shown in the P-V diagram. The net heat absorbed by the gas in this process is (A) (1/2)(P1-P2)(V1-V2) (B) (1/2)(P1+P2)(V1-V2) (C) (1/2)(P1+(a/v21)-P2-(a/v22))(V1-V2) (D) (1/2)(P1+(a/v21)+P2+(a/v22))(V1-V2)

Tardigrade Admin · Accepted Answer

For the cyclic process <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/dec901e31e32362ab4d0c3ac294dc030-.png" /> Heat absorbed = Work done = Area =(1/2)(Δ p) × Δ V =(1/2)(p1-p2) ×(V1-V2) =(1/2)(p1-p2)(V1-V2)

Q. One mole of a vander Waals’ gas obeying the equation
$(P + \frac{a}{V ^{2}}) (V - b) = RT$
undergoes the quasi-static cyclic process which is shown in the $P - V$ diagram. The net heat absorbed by the gas in this process is

$\frac{1}{2} (P_{1} - P_{2}) (V_{1} - V_{2})$

$\frac{1}{2} (P_{1} + P_{2}) (V_{1} - V_{2})$

$\frac{1}{2} (P_{1} + \frac{a}{v _{1}^{2}} - P_{2} - \frac{a}{v _{2}^{2}}) (V_{1} - V_{2})$

$\frac{1}{2} (P_{1} + \frac{a}{v _{1}^{2}} + P_{2} + \frac{a}{v _{2}^{2}}) (V_{1} - V_{2})$

For the cyclic process

Heat absorbed $=$ Work done
$=$ Area $= \frac{1}{2} (Δ p) \times Δ V$
$= \frac{1}{2} (p_{1} - p_{2}) \times (V_{1} - V_{2})$
$= \frac{1}{2} (p_{1} - p_{2}) (V_{1} - V_{2})$

Q. One mole of a vander Waals’ gas obeying the equation (P+V2a​)(V−b)=RT undergoes the quasi-static cyclic process which is shown in the P−V diagram. The net heat absorbed by the gas in this process is

21​(P1​−P2​)(V1​−V2​)

21​(P1​+P2​)(V1​−V2​)

21​(P1​+v12​a​−P2​−v22​a​)(V1​−V2​)

21​(P1​+v12​a​+P2​+v22​a​)(V1​−V2​)