$1.0 \text{ kg}$ of air at pressure $10^6 \text{ Pa}$ and temperature $398 \text{ K}$ expands to a five times greater volume. The expansion occurs such that in every instance the added heat is a quarter of the work done by the gas. Calculate the final pressure. $1 \text{ kmol}$ has a mass of $29 \text{ kg}$ and $C_V = \frac 52 R$.

I solved the problem but I have some questions regarding the approaches I made. I interpreted heat being a quarter of the work done every instance as $\text{d}Q = \frac 14 \text{d}W$. By the 1st law of thermodynamics ($Q = W + \Delta U$), this yields $-\frac 34 \text{d}W = \text{d}U$.

But for molar heat capacity we know that $\text{d}U = nC_V\text{d}T$, hence $-\frac 34 \text{d}W = nC_V\text{d}T$. By definition, $\text{d}W = p\text{d}V$ and therefore $-\frac 34 p\text{d}W = nC_V\text{d}T$. We treat the gas as ideal and thus we can substitute $\frac {nRT}V$ for $p$, arriving at $-\frac 34 \frac{nRT}V \text{d}V = nC_V\text{d}T$.

Separation of variables yield $-\frac 34 \frac{nR}V \text{d}V = \frac {nC_V}T \text{d}T$. We take the definite integrales

$ \displaystyle \int_{V_1}^{5V_1} \!\!\!\!\! -\frac 34 \frac{nR}V \text{d}V = \int_{398}^{T_2} \frac {nC_V}T \text{d}T$

and can then finally solve for $T_2$ (it turns out to be approximately $245 \text{ K}$). At last, we can determine the final pressure from

$\displaystyle \frac {p_1V_1}{T_1} = \frac {p_2V_2}{T_2}$.

(1) My main question is whether you guys know of a different way to solve this problem not involving having to solve differential equations?

(2) I've seen the 1st law written as $\text{d}U = \text{d}Q + \text{d}W$ but if we use that expression we definitely won't get the right answer; what's up with the inconsistency?

(3) I saw no other option than having to use the molar heat capacity equation $\text{d}U = nC_V\text{d}T$ but doesn't this assume constant volume? In our problem the volume is definitely changing, so is it not contradictory to use that very equation?


1 Answer 1


(1)The First law is written in form of differentials themselves,so I think there may be no escape from using differential equations.

(2)The way most commonly the first law is written is,

$dU=dQ-dW_{\text{work done by the system}}$.

Here dW is work done by the system.

However,in subjects other than physics,more important quantity is the work done by the experimenter.(This is quite common in chemistry)

As the process in thermodynamics are most "quasistatic"(http://en.wikipedia.org/wiki/Quasistatic_process),the container/piston is always in equilibrium.

So, $\vec{F_{ext}}=-\vec{F_{int}}$(they are equal and opposite),then we have,

$dW_{\text{by the system}}=-dW_{\text{on the system}}$.

And so the first law can be written as:

$dU=dQ+dW_{\text{work done on the system}}$.

Here dW is work done on the system.

As the two $dW$s have different meanings we won't a different answer.

(3)The internal energy of a gas is state variable/state function(http://en.wikipedia.org/wiki/Functions_of_state).

$U$ depends only on the final and initial states of the system and not on what process was used to get from initial to the final state.So we can use a constant volume process to get $dU$ which then can be used in any process without any modification.


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