The work done on the gas in taking it from D to A is

Question

The work done on the gas in taking it from D to A is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/24f3a4497016531a3dd79d2e17507f88-.png" /> (A) - 414 R (B) + 414 R (C) - 690 R (D) + 690 R

Tardigrade Admin · Accepted Answer

At constant temperature (isothermal process) WDA=nRT In((P1/P2))=2.303×2R×300 log((105/2×105)) =2.303×600R log((1/2)) =0.693×600 R=-414 R.

Q. The work done on the gas in taking it from $D$ to $A$ is

$- 414 R$

$+ 414 R$

$- 690 R$

$+ 690 R$

At constant temperature (isothermal process)
$W_{D A} = n RT I n (\frac{P _{1}}{P _{2}}) = 2.303 \times 2 R \times 300 l o g (\frac{1 0 ^{5}}{2 \times 1 0 ^{5}})$
$= 2.303 \times 600 R l o g (\frac{1}{2})$
$= 0.693 \times 600 R = - 414 R .$

Q. The work done on the gas in taking it from D to A is

−414R

+414R

−690R

+690R