Animated Solution for Physics - Thermodynamics: Half-mole of an ideal monoatomic gas is heated at constant pressure of $1\text{ atm}$ from $20^{\circ }\text{C}$ to $90^{\circ }\text{C}$. Work done by gas is close to (Take, gas constant, $R=8.31\text{ J/mol-K}$)

Imagine you are observing a gas trapped inside a cylinder with a freely moving piston. The problem states that the gas is heated at a constant pressure of $1\text{ atm}$. In thermodynamics, any process that occurs at a constant pressure is called an isobaric process.

When heat is supplied to the gas, it expands, pushing the piston outward. This expansion means the gas is doing work on its surroundings. Our goal is to calculate exactly how much work is done during this expansion as the temperature rises from $20^{\circ }\text{C}$ to $90^{\circ }\text{C}$.

The fundamental formula for the work done by a gas during an isobaric process is simply the pressure multiplied by the change in volume:

However, we hit a roadblock: the problem doesn't tell us the initial or final volumes! But don't panic. Whenever you are missing macroscopic variables like pressure or volume, the Ideal Gas Law is your best friend. The ideal gas equation relates pressure, volume, and temperature:

Since the pressure $p$ is constant throughout this process, any change in volume $\Delta V$ must be directly proportional to a change in temperature $\Delta T$. Mathematically, we can differentiate the ideal gas law at constant pressure to get:

This is a beautiful substitution! We can now replace the $p\Delta V$ term in our work equation with $nR\Delta T$. This gives us a new, incredibly useful formula for work done in an isobaric process when only temperature changes are known:

Before we calculate, let's address a classic trap. The question specifically mentions that it is an "ideal monoatomic gas". Do we actually need to know that it's monoatomic?

Absolutely not! The atomicity of a gas (whether it's monoatomic, diatomic, etc.) determines its degrees of freedom, which is crucial if we were calculating the change in internal energy ($\Delta U=n{C}_v\Delta T$) or the total heat supplied ($\Delta Q=n{C}_p\Delta T$). However, the mechanical work done by expansion ($W=p\Delta V$) is a macroscopic property that applies universally to any ideal gas, regardless of its internal molecular structure. This is a classic distractor designed to make you overthink!

Now, let's plug in our known values into our derived formula. We are given: - Number of moles, $n=\frac{1}{2}$ - Universal gas constant, $R=8.31\text{ J/mol-K}$ - Change in temperature, $\Delta T=90^{\circ }\text{C}-20^{\circ }\text{C}=70^{\circ }\text{C}$

Remember, a change of $1^{\circ }\text{C}$ is exactly equal to a change of $1\text{ K}$. Therefore, a temperature difference of $70^{\circ }\text{C}$ is exactly a difference of $70\text{ K}$.

Substituting these into our equation:

Looking at our options, the closest value is $291\text{ J}$. Therefore, the correct option is (a).