Subjective Type

In an industrial process the volume of $$25.0$$ mol of a monoatomic ideal gas is reduced at a uniform rate from $$0.616\ m^3$$ to $$0.308\ m^3$$ in $$2.00\ h$$ while its temperature is increased at a uniform rate from $$27.0^oC$$ to $$450^oC$$. Through out the process, thegas passes through thermodyamic equilibrium states. What are the cumulative work done on the gas?

Solution

To mole the "uniform rates" described in the problem statement, we have expressed the volume and the temperature functions as follows:
$$V=V_i +\left(\dfrac {V_f -V_i}{\tau_f}\right)t$$ and $$T=T_i +\left(\dfrac {T_f -T_i}{\tau_f}\right)t$$
where $$V_i=0.616\ m^3, V_f=0.308\ m^3, \tau_f=7200\ s, T_i=300\ K$$, and $$T_f=723\ K$$.
We can take to the derivatives of $$V$$ with respect to $$t$$ and use that to evaluate the cumulative work done (from $$t=0$$ unit $$t=\tau$$):
$$W=\displaystyle \int pd\ V=\displaystyle \int \left(\dfrac {nRT}{V}\right)\left(\dfrac {dV}{dt}\right)dt=12.2\ \tau +238113\ \ln (25500- \tau)-2.28\times 10^6$$ with $$SI$$ units understood. With $$\tau =\tau_f$$ our result is $$W=-77169\ J\approx -77.2\ kJ$$, or $$|W|\approx 77.2\ kJ$$.
The graph of cumulative work is shown below. The graph for the work done is purely negative because the gas is being compressed (work is being done on the gas).

SIMILAR QUESTIONS

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