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A rigid container holds $3 \,\text{kg}$ of air. The air is stirred so that its pressure changes from $500 \,\text{kPa}$ to $2000 \,\text{kPa}$, while its temperature is held at $50^\circ\text C$ . The heat transferred is $200 \,\text{kJ}$. Assume the air to behave like an ideal gas with $c_v = 0.714\,\frac{\text{kJ}}{\text{kg}\cdot\text{K}}$. Find the final temperature, the change in internal energy, and the work done.

I am generally confused by what is happening in this question. I imagine we are going from one state to another where the temperature is held constant, and heat is transferred as the pressure rises. Why is it asking for final temperature? Am I miss understanding what is going on?

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  • $\begingroup$ I think you are supposed to assume that that only the initial temperature is 50 C. $\endgroup$
    – Chet Miller
    Sep 29, 2021 at 15:12
  • $\begingroup$ I am still confused ... is this enough information? All 3 variables are changing here, and we need to solve the relation $\Delta U = Q + \Delta W$. The answers are Ans: T=1292.6 K, u= 2076.6 kJ, W=1878.6 kJ, I can't figure it out. $\endgroup$
    – A. Radek Martinez
    Sep 29, 2021 at 15:32
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    $\begingroup$ @A.RadekMartinez Yes, the given information is enough. $\endgroup$
    – Mechanic
    Sep 29, 2021 at 15:45
  • $\begingroup$ Seems like a very unphysical question. Although you can increase the pressure and temp of a gas by "stirring," the idea of quadrupling it's pressure by doing so in a constant volume container is something I've never heard of, and hardly seems possible. Unless your "stirrer" is the exhaust of a rocket nozzle or something. And the work done would be zero if the volume change is zero. I would classify the process as heat addition, or if anything irreversible work $\endgroup$
    – RC_23
    Sep 30, 2021 at 4:49

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So, if I understood the problem correctly, you do work on the ideal gas by stirring it continuously at a constant volume. This increases the temperature of the gas and hence its pressure, as at a constant volume for a constant amount of gas, $P \propto T $. You also transfer $200 \ \text{kJ}$ of heat.

For the work done on the gas to increase the pressure at a constant volume, $T$ cannot be constant, as $P$ is a function of $T$. If $T$ would have been constant, $P$ would have been constant too. I think, as @ChetMiller correctly pointed out in the comments, that the initial temperature only is $50^\circ \text{C}$.

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