Based on the first law of thermodynamics, which one of the following is correct?

Question

Based on the first law of thermodynamics, which one of the following is correct? (A) For an isochoric process, textΔE text= - textq (B) For an adiabatic process, textΔE text= - textw (C) For an isothermal process, textq = textW (D) For a cyclic process, textq = - textW

Tardigrade Admin · Accepted Answer

(a) For isochoric process, textΔV text= text0 textW text= textpΔV text= text0 ∴ textΔE text= textq (b) For adiabatic process, q=0 textΔE text= textW (c) For isothermal process, textΔT text= text0 and textΔE text= text0 textq text= - textW (d) For cyclic process, state functions like textΔE text= text0 textq text= - textW

Q. Based on the first law of thermodynamics, which one of the following is correct?

For an isochoric process, $ΔE = - q$

For an adiabatic process, $ΔE = - w$

For an isothermal process, $q = W$

For a cyclic process, $q = - W$

(a) For isochoric process, $ΔV = 0$

$W = pΔV = 0$

$∴ ΔE = q$

(b) For adiabatic process, $q = 0$

$ΔE = W$

(c) For isothermal process, $ΔT = 0$

and $ΔE = 0$

$q = - W$

(d) For cyclic process, state functions like

$ΔE = 0$

$q = - W$

Q. Based on the first law of thermodynamics, which one of the following is correct?

For an isochoric process, ΔE=−q

For an adiabatic process, ΔE=−w

For an isothermal process, q=W

For a cyclic process, q=−W

(a) For isochoric process, ΔV=0 W=pΔV=0 ∴ΔE=q (b) For adiabatic process, q=0 ΔE=W (c) For isothermal process, ΔT=0 and ΔE=0 q=−W (d) For cyclic process, state functions like ΔE=0 q=−W

For an isochoric process, $ΔE = - q$

For an adiabatic process, $ΔE = - w$

For an isothermal process, $q = W$

For a cyclic process, $q = - W$

(a) For isochoric process, $ΔV = 0$

$W = pΔV = 0$

$∴ ΔE = q$

(b) For adiabatic process, $q = 0$

$ΔE = W$

(c) For isothermal process, $ΔT = 0$

and $ΔE = 0$

$q = - W$

(d) For cyclic process, state functions like

$ΔE = 0$

$q = - W$