When 1 mole of gas is heated at constant volume. Temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct?

Question

When 1 mole of gas is heated at constant volume. Temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct? (A) q = -W = 500 J, Δ U = 0 (B) q = Δ U = 500 J, W= 0 (C) q=W= 500 J, Δ U = 0 (D) Δ U = 0 , q=W= -500 J

Tardigrade Admin · Accepted Answer

We know that, Δ H =Δ E + PΔ V When, Δ V=0, ∴ Δ H=Δ E From the first law of thermodynamics, Δ E = q -W In the given problem, Δ H = 500 J - W = - PΔ V, Δ V= 0 So, Δ E = q = 500 J

Q. When 1 mole of gas is heated at constant volume. Temperature is raised from $298$ to $308 K$ . Heat supplied to the gas is $500 J$ . Then which statement is correct?

$q = - W = 500 J, Δ U = 0$

$q = Δ U = 500 J, W = 0$

$q = W = 500 J, Δ U = 0$

$Δ U = 0, q = W = - 500 J$

We know that, $Δ H = Δ E + P Δ V$
When, $Δ V = 0, ∴ Δ H = Δ E$
From the first law of thermodynamics,
$Δ E = q - W$
In the given problem, $Δ H = 500 J$
$- W = - P Δ V, Δ V = 0$
So, $Δ E = q = 500 J$

Q. When 1 mole of gas is heated at constant volume. Temperature is raised from 298 to 308K. Heat supplied to the gas is 500J. Then which statement is correct?

q=−W=500J,ΔU=0

q=ΔU=500J,W=0

q=W=500J,ΔU=0

ΔU=0,q=W=−500J