# Calculating work done on a gas

1. Oct 13, 2016

### Potatochip911

1. The problem statement, all variables and given/known data
n moles of an ideal gas are placed in a frictionless piston with weight $w_p$ and cross-sectional area $A$. The quantity $\gamma = \frac{c_p}{c_v}$ is a constant, the gas is originally at equilibrium values$(P_i, V_i, \theta_i)$ and the external pressure is taken to be zero. Internal Energy given by $U = nc_v\theta$ Find the irreversible work done if grain of weight $w_g$ is added the following ways:

a) All of the grain is added at once to the top of the piston, the piston oscillates and eventually reaches a new equilibrium at $\theta_i$, show that $W_{on}^{irrev}=nR\theta_i(\frac{w_g}{w_p})$

b) If no heat is allowed in or out of the gas (adiabatic process) and and all of the grain is added at once to the top of the piston show that $W_{on}^{irrev}=\frac{nR\theta_i}{\gamma}(\frac{w_g}{w_p})$

2. Relevant equations

$\delta Q = \Delta U + W$

$W_{on} = - \int_{V_i}^{V_f} PdV$

$PV=nR\theta$

$PV^{\gamma} = constant$

3. The attempt at a solution

For part a) we have $W_{on} = -\int_{V_i}^{V_f}PdV=-\int_{V_i}^{V_f}P_{ext}dV$ and $P_{ext} = \mbox{constant} = \frac{w_g+w_p}{A}$ so

$W_{on}=-\frac{w_g+w_p}{A}\int_{V_i}^{V_f}dV= -\frac{w_g+w_p}{A}(V_f-V_i) \Longleftarrow (1)$

using the ideal gas law $V_f = \frac{nR\theta_i}{P_f}=\frac{nR\theta_i A}{w_g+w_p}$ and $V_i =\frac{nR\theta_i A}{w_p}$ so $V_f - V_i = nR\theta_i A(\frac{w_p}{w_p(w_g+w_p)}-\frac{w_g+w_p}{w_p(w_g+w_p)}) = \frac{nR\theta_i A w_g}{(w_g+w_p)w_p} \Longleftarrow (2)$

Subbing (2) into (1) $W_{on} = -\frac{w_g+w_p}{A}\frac{nR\theta_i A w_g}{w_p(w_g+w_p)}=nR\theta_i ( \frac{w_g}{w_p} )$

For part b) It seems to me like the exact same reasoning applies that $W_{on} = -\int_{V_i}^{V_f} P_{ext}dV$ but this will just produce the exact same result as I got in a), i.e. no $\gamma$ will appear in the denominator.

I'm pretty sure I'm missing a key concept here about it being an adiabatic process but I just can't see how it being adiabatic renders this thought process for calculating the work done incorrect.

Last edited: Oct 13, 2016
2. Oct 13, 2016

### Staff: Mentor

In the 2nd scenario, the internal energy changes because the temperature changes. In this scenario, Q is equal to zero.