Question

5.6L of helium gas at STP is adiabatically compressed to 0.7L. taking the initial temperature to be ${{T}_{1}}$ , the work done in the process is:
(A) $\dfrac{9}{8}R{{T}_{1}}$
(B) $\dfrac{3}{2}R{{T}_{1}}$
(C) $\dfrac{15}{8}R{{T}_{1}}$
(D) $\dfrac{9}{2}R{{T}_{1}}$

Answer 1

Adiabatic process is that in which there is no gain or loss of the heat. In other words, the system is insulated and first, we have to find the value of using the formula as ${{T}_{1}}V_{1}^{\gamma -1}={{T}_{2}}V_{2}^{\gamma -1}$ and then, apply the formula of work done on the system in adiabatic expansion as $\dfrac{nR\Delta T}{\gamma -1}$. Solve it.

Complete step by step solution:
-By the term adiabatic we mean that there is no exchange of heat between the system and the surrounding”. i.e. In an adiabatic expansion, which is a thermodynamic process ,there is no gain or loss of the heat i.e. Q remains zero. The adiabatic process occurs very fastly quickly so there is no time to exchange heat between the system and the surrounding , i.e., the system is well insulated. If the adiabatic expansion system does work, then there is the loss of energy in the form of heat and on the other hand, If the work is done on the system, then the energy of the system will increase.
-Adiabatic expansion is a situation whereby an external work acts upon a system at the expense of utilizing internal energy of the gas and results in lowering the temperature of the molecules of gas. It is related to principles of thermodynamics. The terms adiabatic refers to the process in which heat energy is neither gained nor lost by a system.
the temperature T and volume V relationship from the adiabatic gas equation is as;
$T{{V}^{\gamma -1}}$ = constant ------------(1)
Here, T is the temperature of the gas, V is the volume of the gas and $\gamma$ is the ratio of specific heat capacity at constant pressure( ${{c}_{p}}$) and at constant volume( ${{c}_{v}}$) . i.e.
$\gamma =\dfrac{{{c}_{p}}}{{{c}_{v}}}$
And $\gamma$ has different values of the monatomic , diatomic gases etc. respectively like for Monatomic gases such as the noble gases He, Ne, and Ar , the value of $\gamma$ is 1.664 and for diatomic gases it is = 1.4 etc.
Now, considering the numerical;
${{V}_{1}}$ of helium gas = 5.6 L(given),
${{V}_{2}}$ of helium gas =0.7 (given)
Initial temperature of helium gas= ${{T}_{1}}$ (given)
Since, helium is a monatomic gas , then the value of $\gamma$for monoatomic gases is;
$\gamma$=$\dfrac{5}{3}$
Now, using the equation(1) and then putting all these given values ,we get;
${{T}_{1}}V_{1}^{\gamma -1}={{T}_{2}}V_{2}^{\gamma -1}$
$\begin{aligned} &\implies {{T}_{1}}{{(5.6)}^{\dfrac{5}{3}{{-}^{1}}}}={{T}_{2}}{{(0.7)}^{\dfrac{5}{3}{{-}^{1}}}} \\\ &\implies {{T}_{1}}{{(5.6)}^{\dfrac{5-3}{3}}}={{T}_{2}}{{(0.7)}^{\dfrac{5-3}{3}}} \\\ &\implies {{T}_{1}}{{(5.6)}^{\dfrac{2}{3}}}={{T}_{2}}{{(0.7)}^{\dfrac{2}{3}}} \\\ &\implies {{T}_{1}}={{T}_{2}}{{(\dfrac{0.7}{5.6})}^{\dfrac{2}{3}}} \\\ \end{aligned}$
Therefore, we get ${{T}_{2}}=4{{T}_{1}}$
Work done on the system in the adiabatic equation is given as;
=$\dfrac{nR\Delta T}{\gamma -1}$
Here, n is the no of moles, R is the gas constant.
Since, helium is monoatomic gas , so number of moles=4
${{T}_{2}}=4{{T}_{1}}$
$\gamma =\dfrac{{{c}_{p}}}{{{c}_{v}}}$ =$\dfrac{5}{3}$
Then, work done is

&\implies \dfrac{nR\Delta T}{\gamma -1} \\\ &\implies \dfrac{1\times R({{T}_{2}}-{{T}_{1}})}{\dfrac{5}{3}-1} \\\ &\implies \dfrac{1\times R\times (4{{T}_{1}}-{{T}_{1}})}{\dfrac{5-3}{3}} \\\ &\implies \dfrac{1\times R\times 3{{T}_{1}}}{\dfrac{2}{3}} \\\ &\implies \dfrac{3\times R\times 3{{T}_{1}}}{{}} \\\ &\implies \dfrac{9R{{T}_{1}}}{8} \\\ \end{aligned}$$ Hence, when 5.6L of helium gas at STP is adiabatically compressed to 0.7L. taking the initial temperature to be ${{T}_{1}}$ , the work done in the process is: $\dfrac{9}{8}R{{T}_{1}}$ . **So, option (A) is correct.** **Note:** in the Adiabatic expansion the external work is done on the system at the cost of the internal energy of the gas which causes a decrease in the temperature of the gas and thus, the net energy change in the adiabatic system is zero.