Question

A sample of 2 kg of helium (assumed ideal) is taken through the ABC process and another sample of 2 kg of the same gas is taken through the ADC process. Then the temperature of states A and B is as follows:
(given, R=8.3 joules/mol K).

${{T}_{A}}=120.5K,\text{ }{{T}_{B}}=120.5K$
${{T}_{A}}=241K,\text{ }{{T}_{B}}=241K$
${{T}_{A}}=120.5K,\text{ }{{T}_{B}}=241K$
${{T}_{A}}=241K,\text{ }{{T}_{B}}=482K$

Answer 1

Thermodynamics is a branch of physics that deals with heat, work, and temperature, and their relationship to energy, radiation, and physical properties of matter. An ideal gas is a theoretical gas consisting of several uniformly travelling point particles which are not subject to collision between particles.

Complete answer:
Given information from the question.
Looking over the given diagram you can analyse that from going A to B volume is same but pressure changes , and going from B to C pressure is same but volume changes, while from A to D pressure is same but volume changes and from D to C volume is same but pressure changes .
Mass=2kg
For calculating Molecular formula for determining the molar mass; for obtaining the number of moles, divide the mass of the compound by the molar mass of the compound expressed in grammes.
Where for $He$:
$mole=\dfrac{2\times {{10}^{3}}}{4}$
$=500\text{ }moles\text{ }of\text{ }He\to ABC$
As we have found the moles for the state of ABC similarly going for the state of ADC the mass is the same as the previous one that is 2kg, and the gas which is the same He.
Again $500\,moles\to ADC$
Now we’ll calculate the temperature at the states of A and B
At A:-
${{P}_{A}}=5\times {{10}^{4}}N/{{m}^{2}}$
${{V}_{A}}=10{{m}^{3}}$
${{N}_{A}}=500\text{ }mole$
$PV=nRT$
${{T}_{A}}=\dfrac{PV}{nR}\text{ }$
$=\dfrac{5\times {{10}^{4}}\times 10}{500\times 8.3}$
$=120.5K$
Similarly At B:-
${{P}_{B}}=10\times {{10}^{4}}N/{{m}^{2}}$
${{V}_{B}}=10{{m}^{3}}$
${{n}_{B}}\text{ }=500\text{ }mole$
${{T}_{B}}=\dfrac{PV}{nR}$
$=\dfrac{10\times {{10}^{4}}\times 10}{500\times 8.3}=241K$
Hence, the temperature of states A and B for ${{T}_{A}}=120.5K,\text{ }{{T}_{B}}=241K$.

So, option C is the correct answer.

Note:
$PV~=~nRT$
P = Pressure;
V = the volume of material
n = quantity of material
R = ideal constant gas
T = temperature
It's the same with all the gases. It can also be taken from the theory of microscopic kinetics.
The ideal gas principle is useful in that it obeys the ideal gas law, a condensed state equation.