Question

One mole of an ideal monatomic gas undergoes the process $P = \alpha {T^{\dfrac{1}{2}}}$, where $\alpha $ is a constant. Find the work done by the gas if its temperature increases by $50K$.
A. $407.75\,J$
B. $207.75\,J$
C. $307.75\,J$
D. $107.75\,J$

Answer 1

Hint: Here, we will use the formula of molar specific heat of the gas. Then, we will use the formula for heat supplied to the gas and then we will use the molar specific heat in this formula. Than we will use the formula for calculating the internal energy of the gas. Also, we will use the first law of thermodynamics to calculate the work done by the gas.

Formula used:
The formula used for calculating the molar specific heat of the gas is given below
$C = \dfrac{R}{{\gamma - 1}} + \dfrac{R}{{1 - x}}$
Here, $C$ is the molar specific heat, $\gamma $ is the adiabatic exponent, $R$ is the universal gas constant and $x$ is the polytropic index.
Also, the heat supplied to the gas can be calculated by the following formula
$Q = nC\Delta T$
Here, $Q$ is the heat, $n$ is the number of moles, $C$ is the molar specific heat and $\Delta T$ is the change in temperature.
The formula used for calculating the change in the internal energy of the gas is given below
$\Delta U = \dfrac{{nR\Delta T}}{{\gamma - 1}}$
Here, $\Delta U$ is the internal energy, $n$ is the number of moles, $R$ is the universal gas constant, $\Delta T$ is the change in temperature and $\gamma $ is the adiabatic exponent.

Complete step by step answer:
The process given here is
$P = \alpha {T^{\dfrac{1}{2}}}$, where $\alpha $ is a constant
$ \Rightarrow \,\dfrac{P}{{{T^{\dfrac{1}{2}}}}} = \alpha $
$ \Rightarrow \,P{T^{\dfrac{{ - 1}}{2}}} = \alpha $
Comparing the above equation with $P{T^{\dfrac{m}{{1 - m}}}} = \alpha $
Therefore we get $\dfrac{m}{{1 - m}} = - \dfrac{1}{2}$
$ \Rightarrow \,m = - 1$
Now, we know that the value of adiabatic exponent is $\gamma = 1.67$
The formula used for calculating the molar specific heat of the gas is given below
$C = \dfrac{R}{{\gamma - 1}} + \dfrac{R}{{1 - x}}$
$ \Rightarrow \,C = \dfrac{R}{{1.67 - 1}} + \dfrac{R}{{1 - \left( { - 1} \right)}}$
$ \Rightarrow \,C = \dfrac{R}{{0.67}} + \dfrac{R}{2}$
$ \Rightarrow \,C = \dfrac{{10R}}{{67}} + \dfrac{R}{2}$
$ \Rightarrow \,C = 2R$
Therefore, the molar specific heat of the gas is $2R$.
Now, the increase in temperature for the work done is $\Delta T = 50K$
Now, the heat supplied to the one mole of gas can be calculated by the following formula is given below
$Q = nC\Delta T$
$ \Rightarrow \,Q = \left( 1 \right) \times 2R \times 50$
$ \Rightarrow \,Q = 100R$
Now, the formula used for calculating the change in the internal energy of the gas is given below
$\Delta U = \dfrac{{nR\Delta T}}{{\gamma - 1}}$
$ \Rightarrow \,\Delta U = \dfrac{{1R \times 50}}{{1.67 - 1}}$
$ \Rightarrow \,\Delta U = 75R$
Now, from the first law of thermodynamics, we get
$Q = \Delta U + W$
$ \Rightarrow \,100R = 75R + W$
$ \Rightarrow \,W = 25R$
$ \Rightarrow \,W = 25 \times 8.314$
$ \therefore \,W = 207.75\,J$
Therefore, the work done by the gas if its temperature increases by $50K$ is $207.75\,J$.

Hence, option B is the correct option.

Note:Here, remember that it is important to calculate the molar specific heat so that we can put it into the heat supplied to the gas. Also, it is important to calculate the internal energy of the gas so that we can calculate the work done. Also, heat supplied is put into the work done.