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Question

 The volume of an idea gas is decreased form 5.0L to 2400ml at a constant pressure of 2.0 atm. Calculate the work associated with this process in J. Is work being done on the system or by the system?

 If 2.3 X10-2 atm L of work are done on a 10mL volume of gas at a constant pressure of 4.2 atm, what is the new volume of gas?

 An ideal gas at 15atm and 10L is allowed to expand against a pressure of 2atm at a constant temperature. Calculate the work I units of kJ for the gas expansion. Hint Boyles law.

 Consider a automobile piston, the original volume is 40cm3. If the combuston of this mixuer realese 950J

Answer 1

5.0 L = volume 1 = V1

2.40 L = volume 2 = V2

Pressure = 2.0 atm

work = -P∆V

∆V = change in volume = 2.4 L - 5.0 LL = -2.6 L

work = -(2 atm)(-2.6 L) = +5.2 Latm

Converting to joules (J) we get ...

5.2 Latm x 101.325 J / 1 Latm = 527 J of work done on the system (+ value)

work = -P∆V

work = 2.3x10-2 Latm

P = 4.2 atm

∆V = ?

2.3x10^-2 Latm = -4.2 atm x ∆V

∆V = -0.00548 L = -5.48 mls

New volume = 10.0 ml - 5.48 ml = 4.52 mls

Same idea as above. Boyles law is when the temperature is constant.

P1V1 = P2V2

P1 = 15 atm

V1 = 10 L

P2 = 2 atm

V2 = ?

(15)(10) = (2)(x)

x = 75 L

∆V = 75 L - 10 L = 65 L

work = -P∆V

plug and solve

Same as others above. 40 cm3 = 0.040 L

Plug and solve for ∆V