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Brandon Enright
Brandon Enright
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Volume increase in the system is due to work done by the system. Therefore W is negative using your notation. Think of it this way, work done on the system would push the system inwards, decreasing volume. Therefore a volume increase is work done by the system.

Alternatively you could reason using the formula: $dU = dQ - dW$ (using your notation conventions, were $U$ is internal energy, $W$ is work and $Q$ is heat added to the system)

$dW = PdV.$

Therefore

$dU = dQ - PdV$. (Corrected)

Therefore if $dV$ (change in volume) is positive, $dU$ (change in internal energy) is negative.

Volume increase in the system is due to work done by the system. Therefore W is negative using your notation. Think of it this way, work done on the system would push the system inwards, decreasing volume. Therefore a volume increase is work done by the system.

Alternatively you could reason using the formula: $dU = dQ - dW$ (using your notation conventions, were $U$ is internal energy, $W$ is work and $Q$ is heat added to the system)

$dW = PdV.$

Therefore

$dU = dQ - PdV$. (Corrected)

Therefore if $dV$ (change in volume) is positive, $dU$ (change in internal energy) is negative.

Volume increase in the system is due to work done by the system. Therefore W is negative using your notation. Think of it this way, work done on the system would push the system inwards, decreasing volume. Therefore a volume increase is work done by the system.

Alternatively you could reason using the formula: $dU = dQ - dW$ (using your notation conventions, were $U$ is internal energy, $W$ is work and $Q$ is heat added to the system)

$dW = PdV.$

Therefore

$dU = dQ - PdV$.

Therefore if $dV$ (change in volume) is positive, $dU$ (change in internal energy) is negative.

isn't it obvious?
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edit approved Sep 22, 2015 at 15:06
Community Bot
edit approved Sep 22, 2015 at 15:06
Community Bot
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Volume increase in the system is due to work done by the system. Therefore W is negative using your notation. Think of it this way, work done on the system would push the system inwards, decreasing volume. Therefore a volume increase is work done by the system.

Alternatively you could reason using the formula: $dU = Q - dW$$dU = dQ - dW$ (using your notation conventions, were $U$ is internal energy, $W$ is work and $Q$ is heat added to the system)

$dW = PdV.$

Therefore

$dU = Q - PdV$$dU = dQ - PdV$. (Corrected)

Therefore if $dV$ (change in volume) is positive, $dU$ (change in internal energy) is negative.

Volume increase in the system is due to work done by the system. Therefore W is negative using your notation. Think of it this way, work done on the system would push the system inwards, decreasing volume. Therefore a volume increase is work done by the system.

Alternatively you could reason using the formula: $dU = Q - dW$ (using your notation conventions, were $U$ is internal energy, $W$ is work and $Q$ is heat added to the system)

$dW = PdV.$

Therefore

$dU = Q - PdV$.

Therefore if $dV$ (change in volume) is positive, $dU$ (change in internal energy) is negative.

Volume increase in the system is due to work done by the system. Therefore W is negative using your notation. Think of it this way, work done on the system would push the system inwards, decreasing volume. Therefore a volume increase is work done by the system.

Alternatively you could reason using the formula: $dU = dQ - dW$ (using your notation conventions, were $U$ is internal energy, $W$ is work and $Q$ is heat added to the system)

$dW = PdV.$

Therefore

$dU = dQ - PdV$. (Corrected)

Therefore if $dV$ (change in volume) is positive, $dU$ (change in internal energy) is negative.

some TeX formatting
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Waffle's Crazy Peanut
Waffle's Crazy Peanut
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Volume increase in the system is due to work done by the system. Therefore W is negative using your notation. Think of it this way, work done on the system would push the system inwards, decreasing volume. Therefore a volume increase is work done by the system.

Alternatively you could reason using the formula: dU = Q - dW$dU = Q - dW$ (using your notation conventions, were U$U$ is internal energy, W$W$ is work and Q$Q$ is heat added to the system)

dW = PdV.$dW = PdV.$

Therefore

dU = Q - PdV$dU = Q - PdV$.

Therefore if dV$dV$ (change in volume) is positive, dU$dU$ (change in internal energy) is negative.

Volume increase in the system is due to work done by the system. Therefore W is negative using your notation. Think of it this way, work done on the system would push the system inwards, decreasing volume. Therefore a volume increase is work done by the system.

Alternatively you could reason using the formula: dU = Q - dW (using your notation conventions, were U is internal energy, W is work and Q is heat added to the system)

dW = PdV.

Therefore

dU = Q - PdV.

Therefore if dV (change in volume) is positive, dU (change in internal energy) is negative.

Volume increase in the system is due to work done by the system. Therefore W is negative using your notation. Think of it this way, work done on the system would push the system inwards, decreasing volume. Therefore a volume increase is work done by the system.

Alternatively you could reason using the formula: $dU = Q - dW$ (using your notation conventions, were $U$ is internal energy, $W$ is work and $Q$ is heat added to the system)

$dW = PdV.$

Therefore

$dU = Q - PdV$.

Therefore if $dV$ (change in volume) is positive, $dU$ (change in internal energy) is negative.

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Kenshin
Kenshin
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