Imagine a piston containing a sample of ideal gas in thermal equilibrium with a
ID: 2217949 • Letter: I
Question
Imagine a piston containing a sample of ideal gas in thermal equilibrium with a large water bath. Assume that the piston head is perfectly free to move unless locked in place, and the walls of the piston readily allow the transfer of energy via heat unless wrapped in insulation. The piston head is unlocked and the gas is in an equilibrium state. For each of the actions described below, state whether the work W done by the gas, the heat energy Q transferred to the gas, and the change in the internal energy delta U of the gas are positive (+), negative (-), or zero (0). After each action the piston is reset to its initial equilibrium state. A. Action: Lock the piston head in place. Hold the piston above a very hot flame Enter the signs of W,Q , and delta U. Use + ,- , or 0 separated by commas. For example, if W is positive, is Q negative, and delt U is zero, you would type +,-,0. B. Action: Very slowly push the piston head down. Enter the signs of W,Q , and delta U. Use + ,- , or 0 separated by commas. For example, if W is positive, is Q negative, and delt U is zero, you would type +,-,0. C. Action: Lock the piston head in place. Plunge the piston into very cold water. Enter the signs of W,Q , and delta U. Use + ,- , or 0 separated by commas. For example, if W is positive, is Q negative, and delt U is zero, you would type +,-,0. D. Action: Wrap the piston in insulation. Pull the piston head up. Enter the signs of W,Q , and delta U. Use + ,- , or 0 separated by commas. For example, if W is positive, is Q negative, and delt U is zero, you would type +,-,0.Explanation / Answer
A) piston is locked so it will not move as such there will not be any change in volume... so W= 0 , Q is +ve ( heat transferred to gas), U is +ve as Q-W= U..(0,+,+)
B) piston pushed down Work is done on the system so, W is -ve, Q is +ve , U can be +ve or -ve depending upon Q and W. (-, +, -or+)
C) piston is locked , W=0, cold bath is placed so heat wiil be transfered from the system, Q=-ve and U is -ve (0,-,-)
D) system becomes adiabatic so no heat is transferred, Q=0,... piston moves up, work is done by the system so, w=+ve and U= -ve (+, 0, -)