Question

Calculate Q, w, $\delta$U, $\delta$H for the isothermal reversible expansion of 1 mole of an ideal gas from an initial pressure of 1.0 bar to a final pressure of 0.1 bar at a constant temperature of 273 K.

Answer 1

As we can see that this question is based on the concept of thermodynamics for isothermal reversible expansion.
It happens for an ideal gas at constant temperature with the variation in pressure, here the change in temperature will be equal to zero. By the formula of isothermal reversible expansion, we can calculate the values.

Complete step by step answer:
-Now, in this question, let us have a look at the given values; i.e. n (no. of moles) is 1, P$_1$ (initial pressure) 1 bar, and P$_2$ (final pressure) is 0.1 bar, and T (temperature) is 273 K.

-According to the question, we know it is an isothermal expansion, then as mentioned $\delta$T = 0.
-Thus, we can say that $W=-2.303nRT\log \dfrac{{{V}_{2}}}{{{V}_{1}}}$
-In this process we know temperature is constant, so we can write it as P$_1$V$_1$ = P$_2$V$_2$, or ($\dfrac{{{V}_{2}}}{{{V}_{1}}}=\dfrac{{{P}_{1}}}{{{P}_{2}}}$)
Thus, $W=-2.303nRT\log \dfrac{{{V}_{2}}}{{{V}_{1}}}$, R =8.314 (constant)
Now, we will substitute the values, then $W=-2.303\times 1\times 8.314\times 273\times \log \dfrac{1}{0.01}$
Therefore, W = -5227.01 J
-Now, we will calculate the value of Q using the first law of thermodynamics; i.e.
$\delta$U = Q + W, we have $\delta$U = 0
Thus, Q = -W
Then, Q = +5227.01 J

-Now, we have to calculate the $\delta$H, as we know $\delta$H = $\delta$U + $\delta$ (PV)
$\delta$U = 0, and PV = nRT, if here T is equal to zero,
Then, $\delta$H = 0
-In the last we can conclude that Q (net heat) = +5227.01 J, W (work done) = -5227.01 J, $\delta$U = 0, $\delta$H = 0

Note: Don’t get confused while calculating all the values. In this question, $\delta$ U = 0 because it is an isothermal reversible expansion process, so the internal energy according to the first law of thermodynamics is zero.