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Posted by on Saturday, April 19, 2008 at 1:21pm.

The temperature of a monatomic ideal gas remains constant during a process in which 4500 J of heat flows out of the gas. How much work (including the proper + or - sign) is done on the gas?
J

  • Physics - , Saturday, April 19, 2008 at 5:42pm

    Well, to start - You must compress the gas to get heat out at constant temperature --> work in

    The process is isothermal (constant temp)

    In an ideal gas, the internal energy U depends only on T, so the change in U is zero so
    0 = Q - W
    but Q = -4500 J
    so
    0 = -4500 + W
    W = +4500 = work done BY gas
    so -4500 is work done ON the gas

  • Physics - , Sunday, April 20, 2008 at 4:37am

    I'm not sure what we are doing wrong but the computer is saying -4500 is wrong

  • Sign error - , Sunday, April 20, 2008 at 4:49pm

    so
    0 = -4500 - W ======= NOT + W
    W = -4500 = work done BY gas
    so +4500 is work done ON the gas

    I prefaced this whole thing by telling you why W had to be positive then did not notice my sign error.

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