Question

ideal gas law questions

Movable piston A cylinder contains an ideal gas that is at room temperature. The cylinder is sealed with a piston of mass M and cross-sectional area A that is free to move up or down without friction. No gas can enter or leave the cylinder. The piston is at rest. Atmospheric pressure (i.e. the pressure of the air surrounding the cylinder) is P. a. In the space provided, draw a free-body diagram for the piston. Ideal Make sure the label for each force indicates: • the type of force, • the object on which the force is exerted, and • the object exerting the force. Free-body diagram for piston b. In the space provided, draw an arrow to indicate the direction of the net force on the piston. If the net force is zero, state so explicitly. c. Is the force exerted on the piston by the gas inside the cylinder greater than, less than or equal to the force exerted on the piston by the air outside the cylinder? Explain. Net force on piston Write an equation that relates all the forces on your free-body diagram. (Hint: How are these forces related to the net force?) d. Is the pressure of the gas in the cylinder greater than, less than or equal to atmospheric pressure? Explain. Determine the value of the pressure of the gas in the cylinder in terms of the given quantities (Hint: Which of the forces that act on the piston can you use to find the pressure of the gas?) e. A second cylinder contains a different sample of ideal gas at room temperature, as shown at right. The two cylinders and their pistons are identical Is the pressure of the gas in the second cylinder greater than, less than or equal to the pressure in the cylinder above? If you cannot tell, state so explicitly. Explain. Original cylinder Second cylinder

Answer 1

a)

b)

pressure due to gas outside mg pressure due to gas inside

All the forces act on center of mass

there is no net force acting the piston because it is in equilibrium

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c) on the inside force is greater because we have only single force on the inside which has to balance other two forces (mg and pressure due to gas outside)

P_oA + mg = PA

where P_oA is pressure of gas outside and PA is pressure of gas inside

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d) on the inside pressure is greater. Reason is same as part (c)

P = ( P_oA + mg) / A

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e)

pressure will be same as it does not depend on the gas being used.