# How is pumping "Air" into a bicycle make dW (delta work )negative , when in fact there is a increase in pressure and a slight increase in volume? [closed]

Considering that it's an adiabatic process hence $$\mathrm dQ=0$$, and $$\mathrm dU=\mathrm dW$$. And since the work is done on the system then $$\mathrm dW$$ must be positive. (The answer which is in the textbook is that $$\mathrm dU$$ is positive, while $$\mathrm dW$$ is negative.)

• The book must be using the sign convention that dW is the work done by the sysetem on the surroundings, and -dW is the work done by the surroundings on the system. Oct 2 '20 at 11:11

Depending on context and domain dependent convention, the work done is assumed in 2 different, opposite ways, with opposite signs.

Work done on a system $$\Delta U = Q + W$$, preferred by chemistry

versus

work done by a system $$\Delta U = Q - W$$, historically preferred by physics. It is used by the former way as well recently.

Speculatively said, physics way was defined during creation of theoretical background for thermal machines. Physicists may had been focusing on work done by machines.

Later chemists came, who invented their way, focusing on energy of systems.

As a prevention, it is always good to explicitly mention which way is used, to prevent confusion.

If applied, your pump is doing positive "chemical" work on the system=tyre, but the system=tyre does negative "physical" work on the pump. The textbook takes the convention of physics.

In an important sign convention, preferred in chemistry, work that adds to the internal energy of the system is counted as positive. On the other hand, for historical reasons, an oft-encountered sign convention, preferred in physics, is to consider work done by the system on its surroundings as positive.

Wikipedia - Work_(thermodynamics)

I think the most common error in thermodynamic chemistry is the convention of the equation. Some supposed that the first law equation should be $$U = Q + W$$, the others supposed $$U = Q - W$$. Because of the sign in equation W with respect to $$pdV$$. But all these things do not change the system intrinsic.

There are two categories of scientists on this planet. Some, like you, Bagya Aluthge, consider that all energies (heat $$Q$$, work $$W$$) entering a system are positive, and contribute to increase the internal energy $$U$$, so that $$\Delta U = \Delta Q + \Delta W = \Delta Q - p\Delta V$$. But some engineers consider that the system is a machine which, when heated ($$\Delta Q > 0$$), must produce some work $$\Delta W$$, which is then positive : $$\Delta W = + p\Delta V$$. For those engineers, the internal energy $$U$$ is the part of the heat that is not transformed into work : $$\Delta U = \Delta Q - \Delta W = \Delta Q - p\Delta V$$.

So, both both categories of scientists, the result is the same for $$U$$ : $$\Delta U = \Delta Q - p\Delta V$$. Only the sign of the work $$p\Delta V$$ is different.

• It is rather that physicists, creating background for thermal machines and focusing on work done by machines, came before chemists, who invented their way, focusing on energy of systems. Oct 2 '20 at 8:29

It's a simple matter of energy conservation. Here no heat can flow between surrounding and system. If work is done by the system, where do you think that energy comes from?

When a gas does expansion work, a part of kinetic energy is used up in doing work and average kinetic energy of the gas decrease. I ask you, what is the measure of overall kinetic energy of a gas? It is the temperature. Since kinetic energy decreases, the temperature also decreases. Internal energy for an ideal has depends only on temperature of the gas, so the internal energy decreases when work is done by an ideal gas in adiabatic process

The equation for first law of thermodynamics might look different for physics and chemistry but it is just a matterof sign convention.

For convenience, this equation will always be true: $$Q = \Delta U + \int p dV$$ For adiabatic processes this becomes $$-\Delta U = \int p dV$$ On expansion $$dV$$ increases so $$U$$ decreases. $$dV$$ and $$\Delta U$$ will have opposite signs