One mole of ideal gas whose adiabatic exponent - -43 undergoes process P -200-1V then change in internal energy of gas when volume changes from 2m3 to 4m3 is-

The correct option is C 1200 J P_i=200+12 P_f=200+14 T_i=200.5× 21× R T_f=200.25× 41× R Δ U=1R1/3×(801 401)R=1200J . ...

You visited us 14 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

First Law of Thermodynamics

One mole of i...

Question

One mole of ideal gas whose adiabatic exponent $γ = \frac{4}{3}$ undergoes process P $= 200 + \frac{1}{V}$ then change in internal energy of gas when volume changes from $2 m^{3}$ to $4 m^{3}$ is?

400

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

800

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

1200

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

600

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

Open in App

Solution

The correct option is C $1200$ J
$P_{i} = 200 + \frac{1}{2}$

$P_{f} = 200 + \frac{1}{4}$

$T_{i} = \frac{200.5 \times 2}{1 \times R}$

$T_{f} = \frac{200.25 \times 4}{1 \times R}$

$Δ U = \frac{1 R}{1 / 3} \times \frac{(801 - 401)}{R} = 1200 J$ .

Suggest Corrections

Similar questions

One mole of an ideal gas $(γ = 1.4)$ is adiabatically compressed so that its temperature rises from $27^{o} C t o 35^{o} C$ . The change in the internal energy of the gas is (given R = 80. J / mole / K)

1 g

mole of an ideal gas at STP is subjected to a reversible adiabatic expansion to double its volume. Find the change in internal energy

(γ = 1.4)

Q. In the P-V diagram shown, the gas does

5 J

of work in isothermal process ab and

4 J

in adiabatic process

b c

. What will be the change in internal energy of the gas in straight path

c

a

Q. At

27^{\circ} C

two moles of an ideal monatomic gas occupy a volume

V

. The gas expands adiabatically to a volume

2 V

.
Given

{(\frac{1}{2})}^{\frac{2}{3}} = 0.63

Q. The temperature of one mole of an ideal gas increases from

298 K

308 K

when it absorbs

200 J

of heat at constant volume. The change in the internal energy of the gas is:

Join BYJU'S Learning Program

One mole of ideal gas whose adiabatic exponent γ=43 undergoes process P=200+1V then change in internal energy of gas when volume changes from 2m3 to 4m3 is?

The correct option is C 1200JPi=200+12Pf=200+14Ti=200.5×21×RTf=200.25×41×RΔU=1R1/3×(801−401)R=1200J.

One mole of ideal gas whose adiabatic exponent $γ = \frac{4}{3}$ undergoes process P $= 200 + \frac{1}{V}$ then change in internal energy of gas when volume changes from $2 m^{3}$ to $4 m^{3}$ is?

The correct option is C $1200$ J
$P_{i} = 200 + \frac{1}{2}$

$P_{f} = 200 + \frac{1}{4}$

$T_{i} = \frac{200.5 \times 2}{1 \times R}$

$T_{f} = \frac{200.25 \times 4}{1 \times R}$

$Δ U = \frac{1 R}{1 / 3} \times \frac{(801 - 401)}{R} = 1200 J$ .