One mole of an ideal gas for which C v -3-2 R is heated reversibly at a constant pressure of 1 atm from 25- C to 100- C- The H is-Given- R -1-987 calA- 3-775 calB- 37-256 calC- 3725-6 calD- 372-56 cal

The correct option is D 372.56 calWe know that, H = nbsp;U + PV Given number of moles n=1, So, PV = RT, PV = RT We also know, U = nbsp;C_v× (T_2 T_1) for 1 ...

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

Byju's Answer

Standard XII

Chemistry

Difference between Internal Energy and Enthalpy

One mole of a...

Question

One mole of an ideal gas for which $C_{v} = (\frac{3}{2}) R$ is heated reversibly at a constant pressure of 1 atm from $25^{o} C$ to $100^{o} C$ . The $△ H$ is:
Given: R = 1.987 cal

3.775 cal

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

37.256 cal

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

372.56 cal

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

3725.6 cal

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

Open in App

Solution

The correct option is C $372.56 cal$
We know that,
$△ H$ = $△ U$ + P $△ V$
Given number of moles n=1,
So, PV = RT, P $△ V$ = R $△ T$
We also know,
$△ U$ = $C_{v} \times (T_{2} - T_{1})$ for 1 mole of a gas.
Put up value of $Δ U$ in formula,
$= (\frac{3}{2} \times R \times 75)$ + $(R \times 75)$
$= \frac{5}{2} \times R \times 75$
= 372.56 cal (R = 1.987 cal)

Suggest Corrections

Similar questions

Q. One mole of an ideal gas for which

C_{V} = (3 / 2) R

is heated reversibly at a constant pressure of

1 a t m

from

25^{\circ} C

100^{\circ} C

. The

△ H

is:

1

mol of an ideal gas for which

C_{v} = \frac{3}{2} R

is heated reversibly at a constant pressure of

1

atm from

25^{o} C

100^{o} C

. The

△ H

is:
Given:

R = 1.987 cal

Q. One mole of an ideal Gas

(C_{v} = \frac{3}{2} R)

is heated at constant pressure reversibly of 1 atmosphere from

25^{\circ} C

100^{\circ} C

. Calculate the absolute value of

Δ E

in calories.
(Take R = 2 cal/molK)

Q. One mole of a monoatomic ideal gas is heated at constant pressure from 25

^{o}

C to 300

^{o}

C. Calculate the

Δ

Δ

U, work done and entropy change during the process.

(G i v e n C_{v} = \frac{3}{2} R)

Q. One mole of an ideal monoatomic gas is heated at a constant pressure from

0

100

. Then the change in the internal energy of the gas is:

(Given,

R = 8.32 J {m o l}^{- 1} K^{- 1}

)

Join BYJU'S Learning Program

Explore more

Difference between Internal Energy and Enthalpy

Standard XII Chemistry

Join BYJU'S Learning Program

One mole of an ideal gas for which Cv=(32)R is heated reversibly at a constant pressure of 1 atm from 25oC to 100oC. The △H is: Given: R = 1.987 cal

The correct option is C 372.56 calWe know that, △H = △U + P△V Given number of moles n=1, So, PV = RT, P△V = R△T We also know, △U = Cv×(T2−T1) for 1 mole of a gas. Put up value of ΔU in formula, =(32×R×75) + (R×75) =52×R×75 = 372.56 cal (R = 1.987 cal)

One mole of an ideal gas for which $C_{v} = (\frac{3}{2}) R$ is heated reversibly at a constant pressure of 1 atm from $25^{o} C$ to $100^{o} C$ . The $△ H$ is:
Given: R = 1.987 cal