1 answer

Can you help with Q6&7 please Remove Looking pin 50 cm 1.0 atm 20°C 100 cm...

Question:

Remove Looking pin 50 cm 1.0 atm 20°C 100 cm 3.0 am WWW 1. Start 2. Heat to 3 atm. 3. Remove pin. Continue heating to 100 cm

Question 6 6 pts In step 3, the pin is removed and the gas is allowed to expand. The new temperature and pressure are Kand at

Question 7 In step 4, the pin is reinserted and one of the masses is removed. The pressure Select) the volume Select . and th

can you help with Q6&7 please

Remove Looking pin 50 cm 1.0 atm 20°C 100 cm 3.0 am WWW 1. Start 2. Heat to 3 atm. 3. Remove pin. Continue heating to 100 cm 4. Insert pin. Remove mass The figure shows a sample of N2 gas covered by a movable piston, first in its initial state and then undergoing three processes. The heat capacity at constant volume for Nz is 5/2 R. In step 1, when the pin locking the piston is removed, the piston does not move. The piston has an area of 10 cm². The figure gives the volume, pressure, and temperature of the gas. If there is a vacuum above the piston, what is the combined weight of the piston and masses, in Newtons? Please express all answers to three significant figures.
Question 6 6 pts In step 3, the pin is removed and the gas is allowed to expand. The new temperature and pressure are Kand atm. The amount of work done on the gas is U.
Question 7 In step 4, the pin is reinserted and one of the masses is removed. The pressure Select) the volume Select . and the temperature Select)

Answers

For question 6

Since there is vacuum outside of the piston and the volumen expands until it stops, then the Pressure stabilize to the same as the beggining. That is 1 atm.

Now since the mass keep constat we can use the equation

P1V1 Ti P2V2 T2

Solving for T and usng all the quantities known.

20 * 1* 100 T2 T1P2V2 PAVI 40C = 313.15K 1 * 50

The work done for the system is The Pressure in Pascals (Pa) times the area in meters (m), times the distance that moved in meters (m). To change from atm to Pa we need to multiply by the constant 101300, 1atm=101300Pa:

W = P Ad = latm * 101300 Pa/atm * .001m *.05m 5.065 Joules.

For question 7

Since the pin was reinserted right after step 3 and the mass was removed later, then the conditions didn't change and remain equal to step 3. Then the Pressure, Volume and Temperature remain or keep equal as in Step 3.

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