Question

In the thermodynamic process, helium gas obeys the law, $\text{ T P}{{\text{ }}^{{-2}/{5}\;}}\text{ = constant }$. The heat is given to n moles of He in order to raise the temperature from T to 2T is:
A) $\text{ 8RT }$
B) $\text{ 4RT }$
C) $\text{ 16RT }$
D) Zero

Answer 1

In the reversible adiabatic expansion the pressure and volume are related to the heat capacity constant $\text{ }\gamma \text{ }$. The relation is as shown below,
$\text{ P}{{\text{V}}^{\gamma }}=\text{Constant }$
For monoatomic gas the$\text{ }\gamma \text{ =}\dfrac{5}{3}\text{ }$. In this problem, find out the relation of pressure and temperature in terms of heat capacity and ratio for monoatomic gas.

Complete step by step solution:
We have given that, the helium gas obeys a certain law .According to which temperatures and pressure of a gas are related as,
$\text{ T P}{{\text{ }}^{{-2}/{5}\;}}\text{ = constant }$
The temperature of the helium gas raises from T to 2T
The number of moles of helium is ‘n’
We have to find the heat liberated in terms of temperature T.
Helium $\text{ He }$ is a monatomic gas. Thus the value of heat capacity ratio i.e. $\text{ }\gamma \text{ }$.the $\text{ }\gamma \text{ }$values for an ideal gas can be related to the degree of freedom. Thus for monoatomic gas, the $\text{ }\gamma \text{ }$value is,
$\text{ }\gamma \text{ =}\dfrac{5}{3}\text{ }$
We have to find the type of process. The adiabatic reversible expansion related the pressure and temperature with heat capacity ratio$\text{ }\gamma \text{ }$. The relation is as shown below,
$\text{ P}{{\text{V}}^{\gamma }}=\text{Constant }$
Form ideal gas equation,
$\begin{aligned} & \text{ PV = nRT } \\\ & \therefore \text{V = }\dfrac{\text{nRT }}{\text{P}}\text{ } \\\ \end{aligned}$
Substitute the value of V in the adiabatic reversible equation. We have,
$\begin{aligned} & \text{ P}{{\left( \dfrac{\text{nRT }}{\text{P}}\text{ } \right)}^{\gamma }}=\text{Constant } \\\ & \Rightarrow {{\text{P}}^{\text{1}-\gamma }}\text{ }{{\text{T}}^{\gamma }}\text{ = Constant } \\\ \end{aligned}$
Since the number of moles and gas constant is constant,
On further modification the relation becomes,
$\begin{aligned} & \text{ }{{\text{P}}^{\text{1}-\gamma }}\text{ }{{\text{T}}^{\gamma }}\text{ = Constant} \\\ & \Rightarrow {{\text{P}}^{\dfrac{1-\gamma }{\gamma }}}\text{ T = Constant} \\\ \end{aligned}$
For monatomic helium gas is $\text{ }\gamma \text{ =}\dfrac{5}{3}\text{ }$, on substitution in the equation we have,
$\begin{aligned} & \text{ }{{\text{P}}^{\dfrac{1-\dfrac{5}{3}}{\dfrac{5}{3}}}}\text{ T = Constant} \\\ & \Rightarrow {{\text{P}}^{\dfrac{-\dfrac{2}{3}}{\dfrac{5}{3}}}}\text{ T = Constant} \\\ & \Rightarrow {{\text{P}}^{-\dfrac{2}{3}\times \dfrac{3}{5}}}\text{ T = Constant} \\\ & \therefore {{\text{P}}^{-\dfrac{2}{5}}}\text{ T = Constant} \\\ \end{aligned}$
This $\text{ }{{\text{P}}^{-\dfrac{2}{5}}}\text{ T = Constant }$ is the same as that given in the problem. Thus the helium gas undergoes the adiabatic reversible expansion.
The process is said to be adiabatic if no heat enters or leaves the system during any step of the process.
Thus the heat change during the expansion is equal to zero $\,\text{Q = 0 }$ .

Hence, (D) is the correct option.

Note: A student should remember that the question can be further extended to the determination of the internal energy and work done by the system. To calculate the internal energy at the constant volume using the formula,
$\text{ }\Delta \text{U = n}{{\text{C}}_{\text{v}}}\text{R}\Delta \text{T }$
Where n is the number of moles, R is the gas constant and $\text{ }\Delta \text{T }$ is the temperature difference.
For the above problem, the internal enthalpy change would be,
$\text{ }\Delta \text{U = n}{{\text{C}}_{\text{v}}}\text{R(2T}-\text{T) }\Rightarrow \text{n}{{\text{C}}_{\text{v}}}\text{RT }$
From the first law of thermodynamics, the heat of the system can be determined as,
$\text{ }\Delta \text{U = }-\text{ w }$