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A certain mass of gas is expanded from $\pu{1 L}$ at $\pu{10 atm}$ to $\pu{4L}$ at $\pu{5 atm}$ against a constant external pressure of $\pu{1 atm}$. If initial temperature of gas is $\pu{300 K}$ and heat capacity at constant volume of preocess is $\pu{50 J °C-1}$, find enthalpy change of process.

Here are my thoughts:

$$∆H = nC_p\,\Delta T = nC_V\,\Delta T + p\,\Delta V$$

where $p$ is external pressure.

$$\Delta U = c\,\Delta T$$

with $T_1 = \pu{300 K}$ and $T_2 = \pu{600 K}$ (which I got from gas equation), so

$$\Delta U = \pu{50 J K-1} \times \pu{300 K} = \pu{1.5 kJ}$$

Still I am not getting the correct answer. I believe my methodology is right.

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  • $\begingroup$ Look at the units of the heat capacity - it is not per mole. So your equation needs adjustment, I think. $\endgroup$ – Karsten Theis Apr 13 at 11:13
  • $\begingroup$ No I did it with 50 itself taking it has heat capacity $\endgroup$ – Abin Antony Apr 13 at 12:59
  • $\begingroup$ So why do you have the "n" in your equation? $\endgroup$ – Karsten Theis Apr 13 at 13:17
  • $\begingroup$ I did like this for ∆u=c∆T T1=300K and T2 =600K(which I got from gas equation) ∆u=50 x 300=1.5kJ $\endgroup$ – Abin Antony Apr 13 at 13:22
  • $\begingroup$ So now you know ∆U for going from initial state to 600 K at the same volume. You still have to go to the larger volume and get ∆(PV) to find ∆H. The outside pressure should have no effect on your result, as both ∆H and ∆U are state functions. $\endgroup$ – Karsten Theis Apr 13 at 13:47
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Assume an ideal gas

I think we have to assume an ideal gas, otherwise it is not clear how to get the final temperature of the gas. In the OP's comments, the final temperature is determined as 600 K.

The difference between change in energy and change in enthalpy

Enthalpy is defined as:

$$H = U + PV$$

If both P and V change from state one to state two:

$$H_1 = U_1 + P_1 V_1$$

$$H_2 = U_2 + P_2 V_2$$

we can relate the change in enthalpy to the change in energy by subtracting the two equations:

$$\Delta H = \Delta U + P_2 V_2 - P_1 V_1$$

We could write $\Delta(PV)$ for $P_2 V_2 - P_1 V_1$, but maybe that is confusing.

Change in energy of the gas

In the question, the temperature, the volume and the pressure of the gas all change as we are going from initial to final state. Because energy and enthalpy are state functions, we can decide in which order these changes happen. It makes sense to use steps from initial to final that are easy to calculate. I will change temperature first, then volume (the pressure will change as well, and we will see how that plays into the calculation).

Initial state: V = 1L, T = 300 K

Intermediate state: V = 1L, T = 600 K

Final state: V = 5L, T = 600 K

The change in energy from the initial state to the intermediate state is given by the OP:

$$\Delta U = C_V \Delta T = \pu{1.5 kJ}$$

Taking that intermediate state (V = 1L) to the final state (V = 5L) does not involve any change in energy if the gas is ideal. We could expand into a vacuum, or against pressure. In the latter case, the temperature would drop and we would have to transfer heat into the system to the same amount as the work it did on the surrounding.

So for the entire process, $\Delta U = \pu{1.5 kJ}$.

Change in enthalpy of the gas

$$\Delta H = \Delta U + P_2 V_2 - P_1 V_1 = $$

$$ = \pu{1.5 kJ + 5 atm 4 L - 1 atm 10 L} = $$

$$ = \pu{1.5 kJ + 10 atm L = 1.5 kJ - 0.101 kJ = 1.4 kJ}$$

What about the external pressure?

It does not matter for the calculation because the state of the gas does not depend on it. In other problems, where the volume of the gas is not constrained, it might be important, but here it is not.

You could solve this problem going on another path, and keeping track of volume work and heat transferred to get at the changes in energy (using the first law of thermodynamics). This would be more complicated, though.

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