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SVETLANKA909090 [29]

3 years ago

10

A piston-cylinder device has two stops; a lower set and an upper set, that constrain the cylinder. When the piston is on the low

er stops, the volume is 0.4 liters. When the piston reaches the upper stops, the volume is 0.6 liters. The piston-cylinder initially contains water at 100 kpa and a quality of 20%. The water is heated until it is completely transformed to steam. Additionally, the mass of the piston requires a pressure of 300kpa to raise it. When the piston hits the upper stops:
Determine the final pressure in the cylinder

Determine heat input

Determine work for the overall process

Hint: there are 4 states. Think about the process involved

1 answer:

stich3 [128]3 years ago

6 0

Answer:

2080kJ

Explanation:

m(U4 - U1) = Q4 - W4

If P is less than 300kpa, then V = 0.4L

If P is greater than 300kpa, then V = 0.6L

If P is equal to 300kpa, then 0.4L < V < 0.6L

At 100kpa, vf = 0.001043m/kg and vg = 1.693m³/kg

Level 1

v1 = 0.001043 + 0.2 x 1.693 = 0.33964

m = V1/v1 ⇔ 0.4/0.33964

m = 1.178kg

u1 = 417.36 + 0.2 x 2088.7 = 835.1kJ/kg

Level 3

v3 = 0.6/1.178 = 0.5095 < vG = 0.6058 at P3 is equal to 300kpa

NOTE: Piston does reach upper stops to reach saturated vapor

Level 4

v4 = v3 = 0.5095m³/kg = vG at P4

P4 = 361kpa, u4 = 2550.0kJ/kg

1W4 = 1W2 + 2W3 + 3W4 = 0 + 2W3 + 0

1W4 = P2 (V3 - V2) = 300 x (0.6 - 0.4) = 60kJ

1Q4 = m(u4 -u1) + 1W4 = 1.178(2550.0 - 835.1) + 60 = 2080kJ

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Answer:

Following are the responses to the given choices:

Explanation:

For point a:

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So, $(2.50 \times 10^{-3} mol - 2.49 \times 10^{-3} mol)$ i.e. $1 \times 10^{-5}$ mol of $OH^-$ in excess in total volume $(24.9+25.0) \ mL = 49.9 \ mL$ i.e. concentration of $OH^- = 2 \times 10^{-4}\ M$

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$= 25.1 mL \times 0.100 M\\\\ = \frac{25.1}{1000}\ L \times 0.100 \ M\\\\ = 2.51\times 10^{-3} \ mol$

1 mol $H^+$ neutralizes 1 mol of $OH^-$

So, $(2.51 \times 10^{-3}\ mol - 2.50 \times 10^{-3}\ mol)$ i.e. $1 \times 10^{-5} \ mol \ of\ H^+$ in excess in the total volume of $(25.1+25.0) \ mL = 50.1\ mL$ i.e. concentration of $H^+ = 2 \times 10^{-4}\ M$

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