Lifting a vertical piston

Problem

A diatomic gas ($n = 0{,}5\ \text{mol}$, $f = 5$) is in a vertical cylinder of area $A = 5\cdot 10^{-3}\ \text{m}^2$, closed by a frictionless piston of mass $m_p = 10\ \text{kg}$; above the piston is the atmosphere at $P_\text{atm} = 10^5\ \text{Pa}$. The gas receives $Q = 500\ \text{J}$. By how much does the piston rise? (slow transformation at constant pressure).

Solution

Gas pressure. The piston is in equilibrium: the gas supports the atmosphere plus the weight of the piston. The pressure stays constant: $P_\text{gas} = P_\text{atm} + \frac{m_p\, g}{A} = 10^5 + \frac{10\cdot 9{,}81}{5\cdot 10^{-3}} = 1{,}196\cdot 10^5\ \text{Pa}$

Temperature increase. At constant pressure $Q = n\, C_p\, \Delta T$ with $C_p = \frac{7}{2}R$: $\Delta T = \frac{Q}{n\, C_p} = \frac{500}{0{,}5\cdot \frac{7}{2}\cdot 8{,}31} = 34{,}4\ \text{K}$

Volume increase. $\Delta V = \frac{n R\, \Delta T}{P_\text{gas}} = \frac{0{,}5\cdot 8{,}31\cdot 34{,}4}{1{,}196\cdot 10^5} = 1{,}20\cdot 10^{-3}\ \text{m}^3$

Rise of the piston. $\Delta h = \frac{\Delta V}{A} = \frac{1{,}20\cdot 10^{-3}}{5\cdot 10^{-3}} = 0{,}24\ \text{m}$

$\ev{\Delta h \approx 0{,}24\ \text{m}}$

Links

Topics: Thermodynamics Concepts: First law of thermodynamics · Gravitational potential energy Skills: Bubble energy balance Objects: Piston and cylinder · Ideal gas