Question

At 27°C one mole of an ideal gas is compressed isothermally and reversible from the pressure of 2 atm to 10 atm. The value of ΔE and q are
[ R=2cal]
A.0, -965.84cal
B.-965.84cal, -865.58cal
C.+865.58cal,-865.58cal
D.+965.84cal, +865.58cal

Answer 1

Hint: We should use the first law of thermodynamics in order to findΔE and q as the given gas is ideal. Also, in the given process temperature does not change as it is an isothermal process.
Formula:
Work done by an ideal gas in reversible isothermal process, $W=-2.303nRT{{\log }_{10}}\dfrac{{{P}_{1}}}{{{P}_{2}}}$, where n is the number of moles of given gas, T is the temperature of given gas, ${{P}_{1}}$ and ${{P}_{2}}$ is the initial and final pressure of the gas.$$ $$
$\Delta E=q+W$, where $\Delta E$ is the internal energy of the given gas, q is the heat added or released and W is work done by the gas.
Also, $\Delta E=\dfrac{f}{2}nR\Delta T$ where f is the degree of freedom of given gas, $\Delta T$ is change in temperature of the gas and n is the number of moles of given gas.

Complete Step by Step Solution:
In the above question following are given:$T={{27}^{o}}C=27+273=300K$, R=2cal, ${{P}_{1}}=2atm,{{P}_{2}}=10 atm$ and n=1
 We already know from the first law of thermodynamics that, $\Delta E=q+W$. Now, since the reaction is isothermal so temperature will remain constant throughout the process which means $\Delta T=0$.
So, $$ $\Delta E=\dfrac{f}{2}nR\Delta T=\dfrac{f}{2}nR(0)=0$$
Now, Work Done=$W=-(2.303)(1)(2)(300){{\log }_{10}}\dfrac{2}{10}$ $W=-2.303nRT{{\log }_{10}}\dfrac{{{P}_{1}}}{{{P}_{2}}}$
$W=-(2.303)(1)(2)(300){{\log }_{10}}\dfrac{2}{10}$=+965.84cal

As, we have already found the value of W and $\Delta E$. So, now by applying the first law of thermodynamics we can easily calculate the value of q. On putting $\Delta E$=0 and W=+965.48 in $\Delta E=q+W$.
0=q+965.84 or
Q=-965.84cal
 $\Delta E$



Therefore, from the above calculations we can now give the conclusion that option A is the correct answer.

Note:
It should be noted that the first law of thermodynamics can be applied only if the gas is ideal.
Also, one should know that the isothermal process occurs slowly so that there is enough time to exchange heat between system and surrounding so that the temperature of the system always remains the same.