Question

Which of the following expressions is true for an ideal gas?
A.${\left( {\dfrac{{\partial V}}{{\partial T}}} \right)_P} = 0$
B.${\left( {\dfrac{{\partial P}}{{\partial T}}} \right)_V} = 0$
C.${\left( {\dfrac{{\partial U}}{{\partial V}}} \right)_T} = 0$
D.${\left( {\dfrac{{\partial U}}{{\partial T}}} \right)_V} = 0$

Answer 1

Hint: We have to know that, ideal gas is a theoretical gas whose particles consume irrelevant space and have no co-operations, and which therefore submits to the gas laws precisely. Or on the other hand Ideal gas will be gas which keeps every one of the gas laws at all temperature and pressing factors.

Complete answer:
We have to know that the kinetic hypothesis of gas gives the qualities of an ideal gas. A portion of the attributes are as per the following:
The gas particles are in consistent irregular movement. They travel in an orderly fashion until they impact another particle or the mass of the holder.
There is no fascination or shock between the gas atoms.
The gas particles are point masses with no volume.
Every one of the impacts are flexible. No energy is acquired or lost during the impact.
All gases at a given temperature have a similar normal dynamic energy.
For an ideal gas,
$\Delta U = n{C_V}\Delta T$
(or)
$PV = nRT$
Where,
$P$ is a pressure,
$V$ is a volume,
$n$ is a number of moles,
$R$ is a universal gas constant,
$T$ is a temperature.
Now,
$P\dfrac{{\partial V}}{{\partial T}} = nR\dfrac{{\partial T}}{{\partial T}}$
Rearrange the above expression,
${\left( {\dfrac{{\partial V}}{{\partial T}}} \right)_P} = \dfrac{{nR}}{P}$
Again,
$\dfrac{{\partial P}}{{\partial T}}V = nR\dfrac{{\partial T}}{{\partial T}}$
Then,
${\left( {\dfrac{{\partial P}}{{\partial T}}} \right)_V} = \dfrac{{nR}}{V}$
Again,
$\Delta U = n{C_V}\Delta T$
${\left( {\dfrac{{\partial U}}{{\partial V}}} \right)_T} = \dfrac{{\partial \left( {n{C_V}\Delta T} \right)}}{{\partial V}} = 0$
Therefore,
${\left( {\dfrac{{\partial U}}{{\partial T}}} \right)_T} = \dfrac{{\partial \left( {n{C_V}\Delta T} \right)}}{{\partial T}} = n{C_V}$

Hence, option (C) is correct.

Note:
We have to know that the ideal gas law can be utilized to compute the volume of gases devoured or delivered. The ideal-gas condition as often as possible is utilized to interconvert among volumes and molar sums in compound conditions. Start by changing over the mass of calcium carbonate to moles.