One mole of an ideal diatomic gas ( C v =5 cal ) was transformed from initial 25° C and 1 L to the state when temperature is 100° C and volume 10 L. Then for this process ( R =2 calories / mol / K ) (take calories as unit of energy and kelvin for temp)

Question

One mole of an ideal diatomic gas ( C v =5 cal ) was transformed from initial 25° C and 1 L to the state when temperature is 100° C and volume 10 L. Then for this process ( R =2 calories / mol / K ) (take calories as unit of energy and kelvin for temp) (A) Δ H =525 (B) Δ S =5 ln (373/298)+2 ln 10 (C) Δ E =525 (D) Δ G of the process can not be calculated using given information.

Tardigrade Admin · Accepted Answer

Δ S=n Cv ln ((Tf/Ti))+n R ln ((Vf/Vi)) Δ H = nC p Δ T Δ E = nC v Δ T Δ G =Δ H -Δ( TS )

Q. One mole of an ideal diatomic gas $(C_{v} = 5 c a l)$ was transformed from initial $2 5^{\circ} C$ and $1 L$ to the state when temperature is $10 0^{\circ} C$ and volume $10 L$ . Then for this process $(R = 2$ calories $/ m o l / K)$ (take calories as unit of energy and kelvin for temp)

$Δ H = 525$

$Δ S = 5 ln \frac{373}{298} + 2 ln 10$

$Δ E = 525$

$Δ G$ of the process can not be calculated using given information.

$Δ S = n C_{v} ln (\frac{T _{f}}{T _{i}}) + n R ln (\frac{V _{f}}{V _{i}})$
$Δ H = n C_{p} Δ T$
$Δ E = n C_{v} Δ T$
$Δ G = Δ H - Δ (TS)$

Q. One mole of an ideal diatomic gas (Cv​=5cal) was transformed from initial 25∘C and 1L to the state when temperature is 100∘C and volume 10L. Then for this process (R=2 calories /mol/K) (take calories as unit of energy and kelvin for temp)

ΔH=525

ΔS=5ln298373​+2ln10

ΔE=525

ΔG of the process can not be calculated using given information.