A gaseous system undergoes a change of state from (A) to (B) by any of the given path path I or path II as shown in figure As per path I; Δ q=-400 cal and Δ W=14 cal As per path II, Δ q=-48 cal Therefore work done, Δ W in path (II) is

Question

A gaseous system undergoes a change of state from (A) to (B) by any of the given path path I or path II as shown in figure As per path I; Δ q=-400 cal and Δ W=14 cal As per path II, Δ q=-48 cal <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/13f36fa4191d58bd023aa5a99646a114-.png" /> Therefore work done, Δ W in path (II) is (A) -338 cals (B) -366 cals (C) -434 cals (D) -462 cals

Tardigrade Admin · Accepted Answer

For path I Δ E I =Δ q+Δ W=-400+14=-386 cals For path II, Δ E II =Δ E I =-48+Δ W ∴ Δ W=-386+48=-338 cal

Q. A gaseous system undergoes a change of state from (A) to (B) by any of the given path
path I or path II as shown in figure
As per path I;
$Δ q = - 400 c a l$ and $Δ W = 14 c a l$
As per path II, $Δ q = - 48 c a l$

Therefore work done, $Δ W$ in path (II) is

-338 cals

-366 cals

-434 cals

-462 cals

For path I $Δ E_{I} = Δ q + Δ W = - 400 + 14 = - 386 c a l s$

For path II, $Δ E_{II} = Δ E_{I} = - 48 + Δ W$

$∴ Δ W = - 386 + 48 = - 338 c a l$

Q. A gaseous system undergoes a change of state from (A) to (B) by any of the given path path I or path II as shown in figure As per path I; Δq=−400cal and ΔW=14cal As per path II, Δq=−48cal Therefore work done, ΔW in path (II) is