A system is taken through a cyclic process represented by a circle as shown. The heat absorbed by the system is

Question

A system is taken through a cyclic process represented by a circle as shown. The heat absorbed by the system is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-u7sm2hb7fyxryovm.jpg" /> (A) π × 103 J (B) (π /2) J (C) 4π × 102 J (D) π J

Tardigrade Admin · Accepted Answer

In the cyclic process Δ Q= Work done = Area inside the closed curve. Treat the circle as an ellipse of area Δ Q=(π /4)(P2 - P1) (V2 - V1) ⇒ Δ Q=(π /4)(150 - 50)× (10)3.(40 - 20)× (10)- 6 ⇒ Δ Q=(π /2) J

Q. A system is taken through a cyclic process represented by a circle as shown. The heat absorbed by the system is

$π \times 1 0^{3} J$

$\frac{π}{2} J$

$4 π \times 1 0^{2} J$

$π J$

In the cyclic process
$Δ Q =$ Work done $=$ Area inside the closed curve. Treat the circle as an ellipse of area
$Δ Q = \frac{π}{4} (P_{2} - P_{1}) (V_{2} - V_{1})$
$\Rightarrow Δ Q = \frac{π}{4} (150 - 50) \times (10)^{3} . (40 - 20) \times (10)^{- 6}$
$\Rightarrow Δ Q = \frac{π}{2} J$

Q. A system is taken through a cyclic process represented by a circle as shown. The heat absorbed by the system is

π×103J

2π​J

4π×102J

πJ

In the cyclic process ΔQ= Work done = Area inside the closed curve. Treat the circle as an ellipse of area ΔQ=4π​(P2​−P1​)(V2​−V1​) ⇒ΔQ=4π​(150−50)×(10)3.(40−20)×(10)−6 ⇒ΔQ=2π​J

$π \times 1 0^{3} J$

$\frac{π}{2} J$

$4 π \times 1 0^{2} J$

$π J$

In the cyclic process
$Δ Q =$ Work done $=$ Area inside the closed curve. Treat the circle as an ellipse of area
$Δ Q = \frac{π}{4} (P_{2} - P_{1}) (V_{2} - V_{1})$
$\Rightarrow Δ Q = \frac{π}{4} (150 - 50) \times (10)^{3} . (40 - 20) \times (10)^{- 6}$
$\Rightarrow Δ Q = \frac{π}{2} J$