A gas at pressure 6 ×105 N m-2 and volume 1 m3 and its pressure falls to 4 × 105 N m-2. When its volume is 3 m3. Given that the indicator diagram is a straight line, work done by the system is

Question

A gas at pressure 6 ×105 N m-2 and volume 1 m3 and its pressure falls to 4 × 105 N m-2. When its volume is 3 m3. Given that the indicator diagram is a straight line, work done by the system is (A) 6 × 105 J (B) 3 × 105 J (C) 4 × 105 J (D) 10 × 105 J

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/3a74702c5b588e7f8767d681d69950a0-.png" /> Work done by the system = area under P - V diagram = area of rectangle BCDE + area of ÎABC =4×105×2+(2×105×2/2) W=10×105 J

Q. A gas at pressure $6 \times 1 0^{5} N m^{- 2}$ and volume $1 m^{3}$ and its pressure falls to $4 \times 1 0^{5} N m^{- 2}$ . When its volume is $3 m^{3}$ . Given that the indicator diagram is a straight line, work done by the system is

$6 \times 1 0^{5} J$

$3 \times 1 0^{5} J$

$4 \times 1 0^{5} J$

$10 \times 1 0^{5} J$

Work done by the system
= area under $P - V$ diagram
= area of rectangle $BC D E +$ area of $Δ A BC$
$= 4 \times 1 0^{5} \times 2 + \frac{2 \times 1 0 ^{5} \times 2}{2}$
$W = 10 \times 1 0^{5} J$

Q. A gas at pressure 6×105Nm−2 and volume 1m3 and its pressure falls to 4×105Nm−2. When its volume is 3m3. Given that the indicator diagram is a straight line, work done by the system is

6×105J

3×105J

4×105J

10×105J