One mole of an ideal Gas C v -3-2 R is heated at constant pressure reversibly of 1 atmosphere from 25- C to 100- C - Calculate the absolute value of - E in calories- Take R -2 cal - molK

As, (C_p C_v)=R⇒ C_p=C_v+R ⇒ C_p=32R+R=52R Heat given at constant pressure Δ H=nC_p Δ T=1×52R(373 298) ⇒Δ H = 1 ×52× 2 × 75 = 375 cal Work done in the process = ...

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

Byju's Answer

Standard XII

Chemistry

First Law of Thermodynamics

One mole of a...

Question

One mole of an ideal Gas $(C_{v} = \frac{3}{2} R)$ is heated at constant pressure reversibly of 1 atmosphere from $25^{\circ} C$ to $100^{\circ} C$ . Calculate the absolute value of $Δ E$ in calories.
(Take R = 2 cal/molK)

Open in App

Solution

As, $(C_{p} - C_{v}) = R \Rightarrow C_{p} = C_{v} + R$
$\Rightarrow C_{p} = \frac{3}{2} R + R = \frac{5}{2} R$
Heat given at constant pressure
$Δ H = n C_{p} Δ T = 1 \times \frac{5}{2} R (373 - 298)$
$\Rightarrow Δ H = 1 \times \frac{5}{2} \times 2 \times 75 = 375 cal$
Work done in the process $= - P Δ V$
$= - P (V_{2} - V_{1})$
$= - P (\frac{n R T_{2}}{P} - \frac{n R T_{1}}{P})$
(As, PV = nRT because the process is carried out reversibly)
$= - n R (T_{2} - T_{1})$
$= - 1 \times 2 (373 - 298)$
$= - 150 cal$
From first law of Thermodynamics,
$Δ E = q + W$
$= 375 - 150 = 225 cal$

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:

Q. 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

100^{o} C

. The

△ H

is:
Given: R = 1.987 cal

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. A

27^{\circ} C

one mole of an ideal gas is compressed isothermally and reversible from a pressure of

2 a t m

10 a t m

. The value of

Δ E

and

q

are

(R = 2 c a l)

Q. Calculate

Δ U

and

Δ H

in calories if 1 mole of a monoatomic ideal gas is heated at constant pressure of 1 atm from

25^{o} C

50^{o} C

Join BYJU'S Learning Program

Explore more

First Law of Thermodynamics

Standard XII Chemistry

Join BYJU'S Learning Program

One mole of an ideal Gas (Cv=32R) is heated at constant pressure reversibly of 1 atmosphere from 25∘ C to 100∘ C. Calculate the absolute value of ΔE in calories. (Take R = 2 cal/molK)

One mole of an ideal Gas $(C_{v} = \frac{3}{2} R)$ is heated at constant pressure reversibly of 1 atmosphere from $25^{\circ} C$ to $100^{\circ} C$ . Calculate the absolute value of $Δ E$ in calories.
(Take R = 2 cal/molK)