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Problem in Thermodynamics (2)

by Octavian on Wednesday, February 3, 2010

This is the one I pick from Irodov’s.

Two identical vessels are connected by a tube with a valve letting the gas pass from one vessel into the other if the pressure difference \Delta p \geq 1.10 atm. Initially there was a vacuum in one vessel while the other contained the ideal gas at a temperature t_1 = 27^\circ C and pressure p_1 = 1.00 atm. Then both vessels were heated to a temperature t_2 = 107^\circ C. Up to what value will the pressure in the first vessel (which had vacuum initially) increase?

We denote the volume of each vessel is V, and that \Delta p \geq \Delta p_0 = 1.10 atm. Since it is ideal gas, it satisfies the equation of state pV = nRT. For the second vessel, initially we have the equation p_1 V = n_1 Rt_1, where n_1 is the number of moles contained initially in the second vessel. Suppose after we heat up these two vessels to the temperature t_2, we have n_1' moles in the second vessel and n_2 in the first one, where n_1' + n_2 = n_1, since this two-vessels system is closed. The equations of state for both vessels in temperature t_2 are

p_1' V = n_1' R t_2, for the second vessel

p_2 V = n_2 R t_2, for the first vessel

The difference between p_1' and p_2 is of course \Delta p_0, hence

\Delta p_0 = p_1' - p_2 = \frac{n_1' R t_2}{V} - \frac{n_2 R t_2}{V}

but we also have

n_1' + n_2 = \frac{p_1 V}{R t_1}

and by combining these two equations we will have

p_1 \frac{t_2}{t_1} - 2 \frac{n_2 R t_2}{V} = \Delta p_0

where we know that the second term in the LHS of the equation above is -2p_2. Therefore we have

p_2 = \frac{1}{2} \Big( p_1 \frac{t_2}{t_1} - \Delta p_0 \Big)

which answers the problem.

From → TA on Thermodynamics - Spring 2010, Thermodynamics

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