Question

If gas is compressed under adiabatic conditions (allowing no transfer of heat to or from the surroundings) by application of external pressure, the temperature of the gas would increase. Why?

∆E = q – P∆V where q > 1
∆E = q – P∆V where ∆V < 0
∆E = – P∆V where q = 0
If ∆V is negative in compression, then ∆E decreases while T decreases
If ∆V is negative in compression, then ∆E increases while T increases
Energy is added to the system in the form of heat

Energy is added to the system in the form of work

Answer 1

here is my reasoning:
for an adiabatic trasformation, the first principle of thermodynamic reads:
\[\Delta U = - L = - p\Delta V\]
where \(U\) is the internal energy of the gas, and\(V\) is its volume.
Now, if the gas makes an expansion, then \( \Delta V>0\), so \( \Delta U <0\), according to my formula above.
Next, please keep in mind that the internal energy of a gas is a function of the absolute temperature \(T\): \(U=f(T)\), so, finally, what we get, is a decreasing of absolute temperature \(T\).
Similarly if we compress the gas, this time we have \( \Delta V<0\) and then \(\ Delta U >0\), so we have an increasing of the absolute temperature \(T\)

Answer 2

oops.. thermodynamics*

Answer 3

furthermore, \(p\) is the pressure of the gas

Answer 4

so it would be ∆E = – P∆V where q = 0 ,If ∆V is negative in compression, then ∆E increases while T increases,Energy is added to the system in the form of work?