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

Isothermal process is that in which there is temperature constant. In other words, the system loses heat to the surroundings and we can find the work done in isothermal expansion by using the formula as: $work\text{ }done=2.303\text{ }n\text{ }RT\log \dfrac{{{P}_{1}}}{{{P}_{2}}}$. Now solve it.

Complete step by step solution:
First of all, let’s discuss what is isothermal expansion. By the term isothermal we mean that there is no change in temperature i.e. the temperature of the system remains constant.
In isothermal expansion, which is a thermodynamic process , the gas undergoes expansion at constant temperature and there is loss of the heat to the surroundings so as to maintain the system at constant temperature .i.e. T remains zero. If the isothermal 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.
Now considering the statement as;
We can find the work done during isothermal expansion by simply using the formula of work done in isothermal expansion as;
$work\text{ }done=2.303\text{ }n\text{ }RT\log \dfrac{{{P}_{1}}}{{{P}_{2}}}$ ------------(1)
Here n is the moles of the gas, R is the gas constant, T is the temperature and ${{P}_{1}}$ and ${{P}_{2}}$ are the pressures of the gas.
Now, we know that;
n=1 (given)
R=$8.314$
T=$300K$(given)
${{P}_{1}}=10\text{ }atm$
${{P}_{2}}=1\text{ }atm$
Now put all these values in equation (1), we get;
$\begin{aligned} & work\text{ }done=2.303\times 1\times 8.314\times 300\text{ }\log \dfrac{10}{1}\text{ (}\log 10=1) \\\ & \text{ =2}.303\times 1\times 8.314\times 300\times 1 \\\ & \text{ =5744}\text{.1 J} \\\ \end{aligned}$
So, thus work done during isothermal expansion of one mole of an ideal gas from $10\text{ }atm$ to $1\text{ }atm$ at $300K$ is $5744.1\text{ }J$.

Hence, option (c) is correct.

Note: In isothermal expansion, temperature is constant whereas on the other hand in the adiabatic, heat is constant i.e. process occurs very fastly and quickly so that there is no time to exchange heat between the system and the surrounding , i.e., the system is well insulated and Q=0.