Work and Internal Energy for Adiabatic Processes

[JC2000]

Homework Statement

In an adiabatic expansion, a gas does 25J of work while in an adiabatic compression, 100 J of work **is done on** the gas. Find the change in internal energy for the processes.

Relevant Equations

$dQ = dU + dW$ (where dW is the work done by the gas) ... (1)
$dQ = 0$ for adiabatic processes. ... (2)
$W = \frac {nR(T_1 - T_2)}{\gamma -1}$ (not required but relates to my question)... (3)

Using (2) on (1) give $dU = -dW$... (4)
A.For expansion since the gas goes from $(P_1, V_1, T_1)$ to $(P_2, V_2, T_2)$, does this imply $T_1 \leq T_2$?
B. If so, then $W$ for adiabatic expansion would be negative (using (3))? Using negative $dW$ in (4) gives us a positive result for $dU$?
C. For compression, it would be the opposite?
D. But since the question mentions work done on the gas we can read this as negative work done by the gas, which gives us the same sign as in B.?

 

[JC2000]

Additionally, I ran into the following problem :

In an adiabatic compression, the decrease in volume is associated with : (increase/decrease in temperature and increase/decrease in pressure)? (Sol. : During an adiabatic compression, temperature and **hence internal energy** of the gas increases. In compression pressure will increase.)

E. I understand that since compression occurs, pressure increases, what I don't understand, however, is why temperature increases, how this is deduced. (?)

Thanks!

 

[Chestermiller]

Problem Statement: In an adiabatic expansion, a gas does 25J of work while in an adiabatic compression, 100 J of work **is done on** the gas. Find the change in internal energy for the processes.
Relevant Equations: $dQ = dU + dW$ (where dW is the work done by the gas) ... (1)
$dQ = 0$ for adiabatic processes. ... (2)
$W = \frac {nR(T_1 - T_2)}{\gamma -1}$ (not required but relates to my question)... (3)

Using (2) on (1) give $dU = -dW$... (4)
A.For expansion since the gas goes from $(P_1, V_1, T_1)$ to $(P_2, V_2, T_2)$, does this imply $T_1 \leq T_2$?
B. If so, then $W$ for adiabatic expansion would be negative (using (3))? Using negative $dW$ in (4) gives us a positive result for $dU$?
C. For compression, it would be the opposite?
D. But since the question mentions work done on the gas we can read this as negative work done by the gas, which gives us the same sign as in B.?

The correct form of the first law to apply to this problem is $$\Delta U=Q-W$$ That is because you are dealing with finite changes that may or may not be reversible. For the expansion, the work done by the system on its surroundings is W = +25 J, so $\Delta U=-25 J$. For the compression, the work done by the system on its surroundings is W = -100 J, so $\Delta U = +100 J$.

 

[Chestermiller]

Additionally, I ran into the following problem :

In an adiabatic compression, the decrease in volume is associated with : (increase/decrease in temperature and increase/decrease in pressure)? (Sol. : During an adiabatic compression, temperature and **hence internal energy** of the gas increases. In compression pressure will increase.)

E. I understand that since compression occurs, pressure increases, what I don't understand, however, is why temperature increases, how this is deduced. (?)

Thanks!

The internal energy of an ideal gas increases monotonically with increasing temperature. You have done adiabatic work on the gas (i.e., on the molecules at the boundary of the gas, that then transfer the energy via collisions to the molecules further from the boundary), so their kinetic energy increases, which corresponds to an increase in temperature and internal energy.

 

[JC2000]

The correct form of the first law to apply to this problem is $$\Delta U=Q-W$$ That is because you are dealing with finite changes that may or may not be reversible. For the expansion, the work done by the system on its surroundings is W = +25 J, so $\Delta U=-25 J$. For the compression, the work done by the system on its surroundings is W = -100 J, so $\Delta U = +100 J$.

The internal energy of an ideal gas increases monotonically with increasing temperature. You have done adiabatic work on the gas (i.e., on the molecules at the boundary of the gas, that then transfer the energy via collisions to the molecules further from the boundary), so their kinetic energy increases, which corresponds to an increase in temperature and internal energy.

Thank you! That makes a lot of sense!
Apologies for phrasing my questions quite badly (A,B,C,D).