1. The problem statement, all variables and given/known data
I am wondering about the following conventions:
The work done by the gas is positive if the gas expands,
and negative if it is compressed.
Conversely, the work done on the gas by external force
(e.g., a moving piston) is positive for compression, and negative when the gas expands.
Why is $$ W_{A,B}=-\int_{A}^{B}PdV $$
Specifically the signs are driving me nuts in working out the total work done on the gas in the Carnot cycle.
2. Relevant equations
So if a piston is compressing the work done on the gas is by convention positive. But $$dV$$ is negative. So is the negative sign in $$ W_{A,B}=-\int_{A}^{B}PdV $$
a result of trying to make sure the work done is positive in compression?.
3. The attempt at a solution
Consider the Carnot cycle 1st stage of :
isothermal expansion from volume $$V_{1}$$ to volume $$V_{2}$$ at constant temperature $$T_{1}$$. If I use the above definition I get:
$$W=-\int_{V_{1}}^{V{2}}PdV=nRT_{1}Log\left(\frac{V_{1}}{V_{2}}\right)$$
However in this case $$dV$$ is positive so do I just drop the minus from the integral? And hence is the true answer:
$$W=\int_{V_{1}}^{V{2}}PdV=nRT_{1}Log\left(\frac{V_{2}}{V_{1}}\right)$$
which makes no sense if we define the work on a gas as it expands as negative. ie, the above Log will always be positive. If my first answer is correct this would imply that the total work on the gas in the Carnot cycle is negative, while the work done by the gas is positive. Surely this makes sense as we are converting heat into the external work? Thanks in advance.
I am wondering about the following conventions:
The work done by the gas is positive if the gas expands,
and negative if it is compressed.
Conversely, the work done on the gas by external force
(e.g., a moving piston) is positive for compression, and negative when the gas expands.
Why is $$ W_{A,B}=-\int_{A}^{B}PdV $$
Specifically the signs are driving me nuts in working out the total work done on the gas in the Carnot cycle.
2. Relevant equations
So if a piston is compressing the work done on the gas is by convention positive. But $$dV$$ is negative. So is the negative sign in $$ W_{A,B}=-\int_{A}^{B}PdV $$
a result of trying to make sure the work done is positive in compression?.
3. The attempt at a solution
Consider the Carnot cycle 1st stage of :
isothermal expansion from volume $$V_{1}$$ to volume $$V_{2}$$ at constant temperature $$T_{1}$$. If I use the above definition I get:
$$W=-\int_{V_{1}}^{V{2}}PdV=nRT_{1}Log\left(\frac{V_{1}}{V_{2}}\right)$$
However in this case $$dV$$ is positive so do I just drop the minus from the integral? And hence is the true answer:
$$W=\int_{V_{1}}^{V{2}}PdV=nRT_{1}Log\left(\frac{V_{2}}{V_{1}}\right)$$
which makes no sense if we define the work on a gas as it expands as negative. ie, the above Log will always be positive. If my first answer is correct this would imply that the total work on the gas in the Carnot cycle is negative, while the work done by the gas is positive. Surely this makes sense as we are converting heat into the external work? Thanks in advance.
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