Question

If \[R\] is universal gas constant, the amount of heat needed to raise the temperature of \[2moles\] of an ideal monatomic gas from \[273{\text{ }}K\] to \[373{\text{ }}K\] when no work is done.
A. $100R$
B. $150R$
C. $300R$
D. $500R$

Answer 1

Hint: Let us first obtain a fundamental understanding of the universal gas constant before moving on to the question. The gas constant is a proportionality constant that connects the energy scale in physics to the temperature scale and the scale used to measure substance quantities. As a result, the value of the gas constant is ultimately determined by past decisions and accidents in the definition of energy, temperature, and material amount.

Complete answer:
Now, coming to the question; it given to us that during the process of changing temperature from
\[273{\text{ }}K\] to \[373{\text{ }}K\] no work done happened. Therefore it means that it is an isochoric process.
[The work done in such a process is zero (because \[dW = {\text{ }}P \cdot dV = {\text{ }}0\] while \[V\] is constant). As a result of the first law of thermodynamics, \[dQ = dU\] (isochoric process), the total heat supplied or rejected equals the rise or decrease in the system's internal energy.]
The degrees of freedom of the system are calculated as follows:
 \[f = 3{\text{ }}N-K\]
\[f\] denotes the number of degrees of freedom.
The number of particles in the system is denoted by the letter \[N\] .
\[K\] denotes a particle-to-particle relationship that is not dependent on each other.
For a monatomic gas with \[N = 1\] and \[K = 0\] .
\[f = 3{\text{ }} \times {\text{ }}1-0{\text{ }} = 3\]
Hence, a monatomic gas has three degrees of freedom.
Now, since it is a monatomic gas it’s degree of freedom will be $3$
$\therefore {C_V} = \dfrac{{fR}}{2} = \dfrac{{3R}}{2}$
( \[{C_V}\] denotes the molar heat capacity \[C\] when volume is constant. When a substance has a constant volume, its volume does not change, hence the volume change is zero.)
When it comes to the isochoric process, heat supply is given by;
$Q = n{C_V}\Delta T = 2 \times \dfrac{{3R}}{2} \times \left( {373 - 273} \right) = 300R$
Thus, the amount of heat needed to raise the temperature of \[2moles\] of an ideal monatomic gas from \[273{\text{ }}K\] to \[373{\text{ }}K\] when no work is done is $300R$

So, the correct option is: (C) $300R$

Note:
Students should be aware that for gases with molecules with a higher number of atoms, the degree of freedom will vary much more due to the enormous number of potential arrangements. Thus, for such gases, knowing the degree of freedom can aid in determining the particular molar heats and their ratios, which will obviously decrease as atomicity increases.