You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Entropy Change in Isobaric Process

What is the e...

Question

What is the entropy change in $2 m o l N_{2}$ when its temperature is taken from $400 K$ to $800 K$ , at constant pressure?

30 J / K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

60 J / K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40 J / K

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

20 J / K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $40 J / K$
$Δ S = n C_{P} l n \frac{T_{2}}{T_{1}}$

$C_{v} (for diatomic) = \frac{5}{2} R$
$C_{p} = C_{v} + R$
$C_{P} = \frac{7}{2} R$
$∴$
$Δ S = 2 \times \frac{7}{2} R \times l n \frac{800}{400} = 40.33 \approx 40 J / K$

Suggest Corrections

Similar questions

Q. Calculate the change in entropy when 1 mol of solid iodine at a temperature of 360 K is heated at constant pressure to produce liquid iodine at a temperature of 410 K. The constant pressure molar heat capacity of solid iodine is 54.44 J

K^{- 1}

m o l^{- 1}

and liquid iodine is 80.67 J

K^{- 1}

m o l^{- 1}

.
The melting point of iodine is 387 K and molar enthalpy of fusion of iodine is 7.87 kJ/mol.

Q. When

2 m o l e s

of an ideal monoatomic gas are heated from

300 K

600 K

at constant pressure. The change in entropy of gas

(△ S)

is:

Pressure of 10 moles of an ideal gas is changed from 2 atm to 1 atm against constant external pressure without change in temperature. If surrounding temperature (300 K) and pressure (1 atm) always remains constant then calculate total entropy change ( $Δ S_{s y s t e m} + Δ S_{s u r r o u n d i n g})$ for given process. [Given: ℓn2 = 0.70 amd R=8.0 J/mol/K]

Q. The temperature of 20 L of

N_{2}

was increased from 100 K to 300 K at a constant pressure change in volume:

Q. One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :
(

R = 0.082 L a t m {m o l}^{- 1} K^{- 1} = 8.3 {m o l}^{- 1} K^{- 1}

)

Join BYJU'S Learning Program

What is the entropy change in 2 mol N2 when its temperature is taken from 400 K to 800 K, at constant pressure?

The correct option is C 40 J/KΔS=nCPlnT2T1 Cv (for diatomic) =52R Cp=Cv+R CP=72R ∴ ΔS=2×72R×ln800400=40.33≈40 J/K

What is the entropy change in $2 m o l N_{2}$ when its temperature is taken from $400 K$ to $800 K$ , at constant pressure?

The correct option is C $40 J / K$
$Δ S = n C_{P} l n \frac{T_{2}}{T_{1}}$

$C_{v} (for diatomic) = \frac{5}{2} R$
$C_{p} = C_{v} + R$
$C_{P} = \frac{7}{2} R$
$∴$
$Δ S = 2 \times \frac{7}{2} R \times l n \frac{800}{400} = 40.33 \approx 40 J / K$