A piston-cylinder system initially contains V_1 = 0.05 m^3 of liquid water at T

Question

A piston-cylinder system initially contains V_1 = 0.05 m^3 of liquid water at T = 40 degree C and p = 200 kPa. The system is heated at constant pressure until the entire liquid is vaporized. The final state is thus saturated vapor. Find the mass of water m, the final temperature T_2, and the final volume, V_2. Use the steam tables for H_2O data. A tank of volume V_1 = 1m^3 contains air at T = 283 K and p_1 = 350 kPa The tank is connected through a valve to a second tank that contains m_2 = 3kg of air at T = 308K and p_2 = 200 kPa. The valve is opened and the entire system reaches thermal equilibrium with the surroundings that are at T = 293 K. Find the volume of the second tank, V_2, and the final equilibrium pressure of air, p. Assume that air can be treated as an ideal gas. The molar mass of air is M = 28.97 g/mole and R = 8.314J/moIe K.

Explanation / Answer

3. A piston cylinder system initially contains V1= 0.05m3 of liquid water at T1= 400C and P1= 200kPa

The final temperature T2 is 1000C

We know V1/V2 = T1/T2

V2 = V1 T2 /T1 = 0.05 100/40 = 5/40 = 0.125m3

From steam table, the density at 400C = mass/ volume

Mass of water = volume * density = 0.05m3 * 992.2187 kg/m3 = 49.6 kg

4. For the second tank,

We know V1/V2 = T1/T2

V2 = V1 T2 /T1 = V1 293/308 = 0.95 V1

This tank contains air of mass 3kg, air density is 1.225kg/m3

V1= mass/density = 3/1.225 m3 = 2.45m3

V2 = 2.33 m3

Considering it to be ideal gas, P1V1 = P2V2

Here, 200kPa * 2.45m3 = P * 2.33m3

P = 210kPa

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