Kinetic Theory & Thermodynamics Test

Result

Kinetic Theory & Thermodynamics Test - 1

Score
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Question 1
1 / -0

For an ideal gas:

A
change in internal energy of gas and the work done by the gas are equal in magnitude in an adiabatic process.

B
no heat is added or removed in an adiabatic process.

C
change in internal energy for a constant pressure process from temperature T₁ to T₂ is equal to n C_v ( T₂ – T₁ ) , where C_vis the molar specific heat at constant volume and ‘n’ is the number of moles of gas

D
internal energy does not change in an isothermal process.

Solution

For all thermal processes.
ΔU = n C_v ΔT where ΔT = (T₂ – T₁)

In an isothermal process, ΔT = 0
ΔU = 0

According to first law of thermodynamics,
ΔQ = ΔU + ΔW

For an adiabatic process, ΔQ = 0
0 = ΔU + ΔW
|ΔU| = |ΔW|

The correct answer is: change in internal energy for a constant pressure process from temperature T₁ to T₂ is equal to n C_v ( T₂ – T₁ ) , where C_v is the molar specific heat at constant volume and ‘n’ is the number of moles of gas, change in internal energy of gas and the work done by the gas are equal in magnitude in an adiabatic process., internal energy does not change in an isothermal process., no heat is added or removed in an adiabatic process.
Question 2
1 / -0

For a monoatomic ideal gas undergoing an adiabatic change, relation between temperature and volume is T V^x = constant , where x is

A
7/5

B
1/3

C
2/5

D
2/3

Solution

For an adiabatic process
T V^{γ -}¹ = constant

So x = (γ - 1)
For a monoatomic gas γ = 5/3

So x = 2/3