Thermodynamics Easy Questions and Answers | Page - 9

Thermodynamics

An ideal gas A and a real gas B have their volumes increased from V to 2V under isothermal conditions. The increase in internal energy

1. will be same in both A and B
1. will be zero in both the gases
1. of B will be more than that of A
1. of A will be more than that of B

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Under isothermal conditions, there is no change in internal energy.

Correct Option: B

Under isothermal conditions, there is no change in internal energy.

A diatomic gas initially at 18°C is compressed adiabatically to one eighth of its original volume. The temperature after compression will be

1. 18°C
1. 668.4°K
1. 395.4°C
1. 144°C

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Initial temperature (T₁) = 18°C = 291 K
Let Initial volume (V₁) = V
Final volume (V₂) = V
8

According to adiabatic process, TV^{γ - 1} = constant
According to question, T₁V₁^{γ - 1} = T₂V₂^{γ - 1}
⇒ T₂ = 293 V₁ ^{γ - 1}
V₂

⇒ T₂ = 293(8)^{(7 / 5) - 1} = 293 × 2.297 = 668.4K
For diatomic gas , γ = C_p = 7
C_v 5

Correct Option: B

Initial temperature (T₁) = 18°C = 291 K
Let Initial volume (V₁) = V
Final volume (V₂) = V
8

According to adiabatic process, TV^{γ - 1} = constant
According to question, T₁V₁^{γ - 1} = T₂V₂^{γ - 1}
⇒ T₂ = 293 V₁ ^{γ - 1}
V₂

⇒ T₂ = 293(8)^{(7 / 5) - 1} = 293 × 2.297 = 668.4K
For diatomic gas , γ = C_p = 7
C_v 5

An ideal gas undergoing adiabatic change has the following pressure-temperature relationship

1. P^{γ - 1} T^γ = constant
1. P^γ T^{γ - 1} = constant
1. P^γ T^{1 - γ} = constant
1. P^{1 - γ} T^γ = constant

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We know that in adiabatic process, PV^γ = constant....(1)
From ideal gas equation, we know that
PV = nRT
V = nRT .......(2)
P

Puttingt the value from equation (2) in equation (1),
P nRT ^γ = constant
P

P^{(1 - γ)} T^γ = constant

Correct Option: D

We know that in adiabatic process, PV^γ = constant....(1)
From ideal gas equation, we know that
PV = nRT
V = nRT .......(2)
P

Puttingt the value from equation (2) in equation (1),
P nRT ^γ = constant
P

P^{(1 - γ)} T^γ = constant

A sample of gas expands from volume V₁ to V₂. The amount of work done by the gas is greatest, when the expansion is

1. adiabatic
1. isobaric
1. isothermal
1. equal in all cases

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In thermodynamics for same change in volume, the work done is maximum in isobaric process because in P – V graph, area enclosed by curve and volume axis is maximum in isobaric process.
So, the choice (b) is correct.

Correct Option: B

In thermodynamics for same change in volume, the work done is maximum in isobaric process because in P – V graph, area enclosed by curve and volume axis is maximum in isobaric process.
So, the choice (b) is correct.

If the ratio of specific heat of a gas at constant pressure to that at constant volume is γ, the change in internal energy of a mass of gas, when the volume changes from V to 2V at constant pressure P, is

1. R
  (γ - 1)
1. PV
1. PV
  (γ - 1)
1. γ PV
  (γ - 1)

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Change in internal energy is equal to work done in adiabatic system
∆W = –∆U (Expansion in the system)
= - 1 (P₁V₁ - P₂V₂)
γ - 1

∆U = 1 (P₂V₂ - P₁V₁)
γ - 1

Here , V₁ = V , V₂ = 2 V
∴ ∆U = 1 [P × 2V - PV] = PV
1 - γ 1 - γ

⇒ ∆U = - PV
γ - 1

Correct Option: C

Change in internal energy is equal to work done in adiabatic system
∆W = –∆U (Expansion in the system)
= - 1 (P₁V₁ - P₂V₂)
γ - 1

∆U = 1 (P₂V₂ - P₁V₁)
γ - 1

Here , V₁ = V , V₂ = 2 V
∴ ∆U = 1 [P × 2V - PV] = PV
1 - γ 1 - γ

⇒ ∆U = - PV
γ - 1