Calculating work done on a gas

[Potatochip911]

Homework Statement

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 $W_{on}^{irrev}=\frac{nR\theta_i}{\gamma}(\frac{w_g}{w_p})$

Homework Equations

$\delta Q = \Delta U + W$

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

$PV=nR\theta$

Adiabatic Process:

$PV^{\gamma} = constant$

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.

 

[Chestermiller]

Homework Statement

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 $W_{on}^{irrev}=\frac{nR\theta_i}{\gamma}(\frac{w_g}{w_p})$

Homework Equations

$\delta Q = \Delta U + W$

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

$PV=nR\theta$

Adiabatic Process:

$PV^{\gamma} = constant$

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.

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