Heating under the piston

Problem

$10\ \text{mol}$ of nitrogen are in a cylinder with a freely sliding piston; the atmosphere at $P_\text{atm} = 1{,}013\cdot 10^5\ \text{Pa}$ presses on the piston. The gas goes from $T_A = 300\ \text{K}$ to $T_B = 320\ \text{K}$. How much heat is needed?

Solution

Volumes from the gas law. At constant pressure $P_\text{atm}$: $V_A = \frac{n R T_A}{P_\text{atm}} = \frac{10\cdot 8{,}31\cdot 300}{1{,}013\cdot 10^5} = 0{,}246\ \text{m}^3$ $V_B = \frac{n R T_B}{P_\text{atm}} = \frac{10\cdot 8{,}31\cdot 320}{1{,}013\cdot 10^5} = 0{,}263\ \text{m}^3$ $\Delta V_\text{gas} = V_B - V_A = 0{,}017\ \text{m}^3$

Work of the atmosphere. The gas expands, so the atmosphere is compressed: $\Delta V_\text{atm} = -\Delta V_\text{gas} = -0{,}017\ \text{m}^3$. $L_\text{atm} = P_\text{atm}\,\Delta V_\text{atm} = 1{,}013\cdot 10^5\cdot(-0{,}017) = -1728\ \text{J}$

First law. The gas itself has no moving walls of its own ($L_\text{gas} = 0$ in our convention), and $\Delta E_\text{int} = \frac{5}{2}\,n R\,\Delta T$: $\Delta E_\text{int} = \frac{5}{2}\cdot 10\cdot 8{,}31\cdot 20 = 4155\ \text{J}$ With $\Delta E_\text{int} = Q + L_\text{atm}$: $4155 = Q - 1728 \ \Rightarrow\ Q = 4155 + 1728 = 5883\ \text{J}$

$\ev{Q \approx 5883\ \text{J}}$

More heat is needed than at constant volume: part increases the internal energy, part pushes the atmosphere away.

Links

Topics: Thermodynamics Concepts: First law of thermodynamics · Internal energy · Ideal gas law Objects: Piston and cylinder · Ideal gas