Question

)You have a box that is rigid on the sides and bottom, but has a piston on the top. There is also a valve attached that can be used to let gas in and out of the box. Initially, the box has 0.15 moles of gas in it and the piston is set so the volume is 3.75L The steps occur sequentially a If the temperature is 20°C, what is the air pressure? b. If a compressor is attached to the valve and another 0.35 moles of gas are introduced, what is the new volume? A weight is put on top of the piston that raises the external pressure to 1.5 atm, what c. volume does the piston compress to? d. If you lock the piston in place after c and open the valve, how much gas will remain in the box?

Answer 1

(a) Assuming gas behave ideally,

PV = nRT (T = 293.15 K)

P = nRT / V = 0.15 moles * 0.0821 L-atm/mol-K *293.15 K / 3.75 L  

P = 0.963 atm.

(b) new volume ?

n = 0.15 + 0.35 moles = 0.5 moles

P = 0.963 atm

V = nRT / P = 0.50 moles * 0.0821 L-atm/mol-K *293.15 K / 0.963 atm  

V = 12.5 L

(c) External P raised to 1.5 atm

we have, P1 V1 = P2 V2

V2 = 0.963 atm*12.5 L / 1.5 atm = 8.025 L

Volume Compressed by weight on piston : 12.5 L -8.025 L = 4.475 L

(d) If valve is opened gas will move out till pressure become equal to 1 atm (outside pressure) :

V = 8.025 L ; P = 1 atm n = ?

n = PV /RT = 1 atm*8.025 L /  0.0821 L-atm/mol-K *293.15 K  

n = 0.33 moles .

(e) T is now = -15 C = 258.15 K

New V = ?

V = nRT / P = 0.33 moles * 0.0821 L-atm/mol-K *258.15 K / 1 atm  

V = 7.07 L

(you can use V1/T1 = V2/T2 also)

(f) change in V =  8.025 L - 7.07 L = 0.958 L

% change = 11.94 %

(we are using Absolute T than C , so this change is as expected )