Question

One mole of an ideal monatomic gas at temperature ${T_0}$ expands solely according to the law $\dfrac{P}{V}$= constant. If the final temperature is $2{T_0}$, heat supplied to the gas is:
(A)$2R{T_0}$
(B) $R{T_0}$
(C) $\dfrac{3}{2}R{T_0}$
(D) $\dfrac{1}{2}R{T_0}$

Answer 1

The first law of thermodynamics gives the mathematical relationship between the internal energy, heat, and work of a system. A monatomic gas is a gas that is made of monatomic particles i.e. the gas is formed of an unreacted single atom, for example, He, Ar, Ne, etc.
Formula used:
$Q = W + \Delta U$
where $Q$ is the heat given to the system, $W$ is the work done by the system$\Delta U$ is the change in internal energy
$\Delta U = \dfrac{f}{2}nR\Delta T$
$f$ is the degree of freedom of the gas molecule,$n$is the number of moles of gas, $R$ is the real gas constant, and $\Delta T$ is the change in temperature of the gas.
$W = \int {PdV}$
$P$ is the pressure of gas and $dV$ is the elemental change in the volume of gas.

Complete step by step solution:
It is given that $\dfrac{P}{V}$ is constant, let the constant value be $C$.
$P$ is the pressure of gas and $V$ is the volume of gas.
$\Rightarrow P = CV$
We know that $W = \int {PdV}$
$W$ is the work done by the system and $dV$ is the elemental change in the volume of gas.
Hence,
${V_2}$ is the final volume of gas and${V_1}$ is the initial volume of gas.
$\Rightarrow W = \int\limits_{{V_1}}^{{V_2}} {CVdV}$
$\Rightarrow W = C\int\limits_{{V_1}}^{{V_2}} {VdV}$
$\Rightarrow W = C[\dfrac{{{V^2}}}{2}]_{{V_1}}^{{V_2}}$
$\Rightarrow W = C[\dfrac{{{V_2}^2}}{2} - \dfrac{{{V_1}^2}}{2}]$
$\Rightarrow W = \dfrac{{C{V_2}^2}}{2} - \dfrac{{C{V_1}^2}}{2}$
We know that $P = CV$
${P_2}$is the final pressure of gas and${P_1}$ is the initial pressure of the gas.
$\Rightarrow W = \dfrac{{{P_2}{V_2}}}{2} - \dfrac{{{P_1}{V_1}}}{2}$
We know that $PV = nRT$
Where $n$ is the number of moles of gas, $R$ is the real gas constant and $T$is the temperature of the gas.
${T_2}$ is the final pressure of gas and${T_1}$ is the initial pressure of the gas.
$\Rightarrow W = \dfrac{{nR{T_2}}}{2} - \dfrac{{nR{T_1}}}{2}$
$\Rightarrow W = \dfrac{{nR({T_2} - {T_1})}}{2}$
$\Rightarrow W = \dfrac{{nR\Delta T}}{2}$
We know that, $\Delta U = \dfrac{f}{2}nR\Delta T$
Where$\Delta U$is the change in internal energy and$f$ is the degree of freedom of the gas molecule.
For monoatomic gas $f = 3$and in question it is given that $n = 1$
Hence,
$\Rightarrow \Delta U = \dfrac{f}{2}nR\Delta T$
$\Rightarrow \Delta U = \dfrac{3}{2}R\Delta T$
And $\Rightarrow W = \dfrac{{R\Delta T}}{2}$
From first law of thermodynamics,
$Q = W + \Delta U$
where $Q$ is heat given to the system.
Hence,
$\Rightarrow Q = \dfrac{{R\Delta T}}{2} + \dfrac{{3R\Delta T}}{2}$
$\Rightarrow Q = 2R\Delta T$

Therefore, the answer to our question is (A)$2R{T_0}$

Note:
One can also directly use the formula to calculate work in a polyprotic process which is $W = \dfrac{{nR\Delta T}}{{1 - k}}$ for a polyprotic process $P{V^k}$is constant. This formula can also be written as $W = \dfrac{{{P_2}{V_2} - {P_1}{V_1}}}{{1 - k}}$. Here the meaning abbreviations used are the same as mentioned above in step-by-step solution.