A gas undergoes the process shown in the diagram below. During the process AB, t

Question

A gas undergoes the process shown in the diagram below. During the process AB, the internal energy of the gas decreases and a certain amount of heat Q goes out of the system for the process CA. Use this information to answer the questions below.

(a) What are the signs of W (work done by the gas), Q, and ?U for the process CA?

(b) What are the signs of W (work done by the gas), Q, and ?U for the process AB?

(c) What are the signs of W (work done by the gas), Q, and ?U for the process BC?

W     ---Select--- positive negative zero Q     ---Select--- positive negative zero ?U     ---Select--- positive negative zero A gas undergoes the process shown in the diagram below. During the process AB, the internal energy of the gas decreases and a certain amount of heat Q goes out of the system for the process CA. Use this information to answer the questions below. A gas undergoes the process shown in the diagram b (a) What are the signs of W (work done by the gas), Q, and ?U for the process CA? W ---Select--- positive negative zero Q ---Select--- positive negative zero ?U ---Select--- positive negative zero (b) What are the signs of W (work done by the gas), Q, and ?U for the process AB? W ---Select--- positive negative zero Q ---Select--- positive negative zero ?U ---Select--- positive negative zero (c) What are the signs of W (work done by the gas), Q, and ?U for the process BC? W ---Select--- positive negative zero Q ---Select--- positive negative zero ?U ---Select--- positive negative zero

Explanation / Answer

Given that for process AB = dU ( internal energy) decreases.

for process CA = dQ ( heat) goes out of system)

a)      For Process CA :

Since Heat is goes out of system                     ( By sign conventions: When heat is given to system = dQ = +Ve

So dQ = -Ve                                                              When heat is taken out of system, dQ = - Ve)

dW ( work) = P (pressure)*dV (change in volume)

since for curve CA , volume is constant . So dV = 0

dW = P*dV

dW = P*0 = 0                     

from first law of thermodynamics, dQ = dU + dW                           ------------------------1

here dQ = heat ,   dU = internal energy, dW = work done

dU = dQ - dW                                                         ( from eq.1)

     = -Ve - 0

dU = -Ve

hence for process CA ,   W = 0, Q = - Ve, dU = -Ve

b) for process AB:

given that dU = decreases.

so by sign conventions , dU = -Ve

dW = P* dV

dW = P* ( V2- V1 )                                      ( for AB volume is decreasing , so V2 - V1 = -ve)

dW    = - Ve

from first law , dQ = dU + dW

                           = -Ve - Ve

                   dQ = - Ve

hence for AB,   W = -Ve,   Q = - Ve , dU = -Ve

c) For process BC:

Since Pressure(P), and volume(V) is increasing for process BC. So

dW = P* dV

dW      = +Ve                                            ( because P & V both are increasing)

Now, PV is directly proportional to T ( Temp.)                                       ( from PV = nR T)

P*V are increasing ,So T ( temp.) will also increase.

since Temp. is increasing, so dU = + Ve

From first law,    dQ = dU + dW

                               = + ve + Ve

                         dQ = + Ve

hence for process BC, W = +ve, Q = + ve, dU = +ve

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