An ideal gas is enclosed in a cylinder with a movable piston on top of it. The p

Question

An ideal gas is enclosed in a cylinder with a movable piston on top of it. The piston has a mass of 8,000 g and an area of 5.00 cm2 and is free to slide up and down, keeping the pressure of the gas constant.

(a) How much work is done on the gas as the temperature of 0.230 mol of the gas is raised from 15.0°C to 250°C?

______ J

My ANSWER that i got wrong --> 1.46e3 J Incorrect: Your answer is incorrect.

(b) What does the sign of your answer to part (a) indicate?

(a)There is no work done, by the gas or the surroundings.

(b)The gas does positive work on its surroundings.

A diatomic ideal gas undergoes the thermodynamic process shown in the PV diagram of the figure below. Determine whether each of the values U, Q, and W for the gas is positive, negative, or zero. Hint: The internal energy of a diatomic ideal gas at pressure P and occupying volume V is given by U = 5/2 PV

.

U     ---Select--- positive, negative or zero Q     ---Select--- positive, negative or zero W     ---Select--- positive, negative or zero

Explanation / Answer

1.

a) Use equation,

W= n*R*deltaT = n*R*(Tf- Ti) = 0.23*8.314*(250-15) = 449 J

b) Since pressure is constant and temperature is increased, volume of the gas increases means pistons goes up.

Hence The gas does positive work on its surroundings.

2. Since the volume decreases work done

W= p*deltaV

W must be negative.

Also by equation,

PV=nRT

we will get Tinitial = Tfinal

means no heat added ot given hence deltaQ =0

By first law of thermodynamics

deltaQ = deltaU + W

we can write,

deltaQ = deltaU - W

Gives deltaU = W

Thus deltaU is negative.

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