The equation of state for a certain gas under isothermal conditions is PV = 31.2, where the units are SI. The work done by this gas as its volume increases isothermally from 1 L to 10 L is approximately
A) 13.6 J.
B) 31.2 J.
C) 71.8 J.
D) 281 J.
E) 312 J.

Question

Quizplus · Accepted Answer

Since the process is isothermal ($\Delta T = 0$), the change in internal energy is also zero $(\Delta U = Q - W \Rightarrow \Delta U = 0 - W = -W)$. Thus, the work done by the gas is equal to the heat absorbed by the gas from the surroundings $(W = -Q)$.

To calculate the heat absorbed, we can use the formula $Q = nRT \ln \left(\frac{V_f}{V_i}\right)$, where $n$ is the number of moles of the gas, $R$ is the gas constant, $T$ is the temperature, $V_i$ is the initial volume, and $V_f$ is the final volume.

We can solve for $n$ from the given equation of state: $PV = nRT \Rightarrow n = \frac{PV}{RT}$. Substituting this into the formula for $Q$, we get:

$Q = \frac{PV}{RT} \cdot RT \ln \left(\frac{V_f}{V_i}\right) = PV \ln \left(\frac{V_f}{V_i}\right)$

Plugging in the given values, we get:

$Q = (31.2\ \mathrm{Pa\cdot m^3})\ln \left(\frac{10\ \mathrm{L}}{1\ \mathrm{L}}\right) \approx 71.8\ \mathrm{J}$

Therefore, the work done by the gas is approximately $-Q = \boxed{\mathrm{(C)\ 71.8\ J}}$.

The Equation of State for a Certain Gas Under Isothermal