The amount of work done in an adiabatic expansion from temperature T to T1 is

Question

The amount of work done in an adiabatic expansion from temperature T to T1 is (A) n R(T-T1) (B) n (R/&gamma;-1)(T-T1) (C) n R T (D) n R(T-T1)(&gamma;-1)

Tardigrade Admin · Accepted Answer

Δ Q=Δ U+Δ W, but Δ Q=0 for adiabatic process. ⇒ Δ W=-Δ U=-n Cν(T1-T)=(n R(T-T1)/&gamma;-1)

Q. The amount of work done in an adiabatic expansion from temperature $T$ to $T_{1}$ is

$n R (T - T_{1})$

$n \frac{R}{γ - 1} (T - T_{1})$

$n RT$

$n R (T - T_{1}) (γ - 1)$

$Δ Q = Δ U + Δ W$ , but $Δ Q = 0$ for adiabatic process.
$\Rightarrow Δ W = - Δ U = - n C_{ν} (T_{1} - T) = \frac{n R ( T - T _{1} )}{γ - 1}$

Q. The amount of work done in an adiabatic expansion from temperature T to T1​ is

nR(T−T1​)

nγ−1R​(T−T1​)

nRT

nR(T−T1​)(γ−1)