This question is really basic. I just want toclear it up. I know dE = q - w. The

Question

This question is really basic. I just want toclear it up. I know dE = q - w. The question states that I loose210 J of heat. Would that make q negative when I plug in 210 to q?It also says that there is constant pressure, 250 Pa, and the gascompresses from .8m^3 to .2m^3. By knowing w = pdV , would the workbe negative since dV would be (.2 - .8)m^3?

By plugging in all of my info, I get the following equal to dE;

dE = (-210) - [250(-.6)] = -60 J or |dE| = 60 J

Is this a correct way of solving the problem? Thanks. This question is really basic. I just want toclear it up. I know dE = q - w. The question states that I loose210 J of heat. Would that make q negative when I plug in 210 to q?It also says that there is constant pressure, 250 Pa, and the gascompresses from .8m^3 to .2m^3. By knowing w = pdV , would the workbe negative since dV would be (.2 - .8)m^3?

By plugging in all of my info, I get the following equal to dE;

dE = (-210) - [250(-.6)] = -60 J or |dE| = 60 J

Is this a correct way of solving the problem? Thanks.

Explanation / Answer

Yes... youre on the right track. . Q is negative if the gas loses heat, so in thiscase      Q = -210 Joules. . The first law says       change in internal energy = Q - W . Work done at constant pressure is just    Q = p(Vf - Vi )    which in yourcase is: .      W = 250 * (0.2 - 0.8) =    -150 Joules . So...     change in internal energy =   -210   - (-150) =    -60Joules    . It is important to NOT absolute value this! The negativechange in internal energy means that the energy decreased, and sothe temperature of the gas also decreased.   But, yes...you have the right ideas for how to do this problem.

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