An ideal gas undergoes a change of state (2.0 atm. 3.0 L, 95 K ) to (4.0 atm. 5.0 L, 245 K) with a change in internal energy, #DeltaU# = 30.0 L atm. The change in enthalpy (#DeltaH#) of the process in L atm is
(A) 44
(B) 42.3
(C) ?

Question

Why don't we use the formula for work as nr(change in temp.) to get the work done by taking n=1.
When to use (change in pv) and when to use(change in temp.)

Wyatt Atkins · Accepted Answer

Well, every natural variable has changed, and so the mols changed as well. Apparently, the starting mols is not #1#!

#"1 mol gas" stackrel(?" ")(=) (P_1V_1)/(RT_1) = ("2.0 atm" cdot "3.0 L")/("0.082057 L"cdot"atm/mol"cdot"K"cdot"95 K")#

#= "0.770 mols" ne "1 mol"#

The final state also presents the same problem:

#"1 mol gas" stackrel(?" ")(=) (P_2V_2)/(RT_2) = ("4.0 atm" cdot "5.0 L")/("0.082057 L"cdot"atm/mol"cdot"K"cdot"245 K")#

#= "0.995 mols" ~~ "1 mol"#

It is clear that with these numbers (did you copy down the question correctly?), the mols of gas changed. So #Delta(nRT) ne nRDeltaT#.

Instead, we begin with the definition:

#H = U + PV#

where #H# is enthalpy, #U# is internal energy, and #P# and #V# are pressure and volume.

For a change in state,

#color(blue)(DeltaH) = DeltaU + Delta(PV)#

#= DeltaU + P_2V_2 - P_1V_1#

#= "30.0 L"cdot"atm" + ("4.0 atm" cdot "5.0 L" - "2.0 atm" cdot "3.0 L")#

#= color(blue)("44.0 L"cdot"atm")#

Had we chosen to use #Delta(nRT)#, we would still get it, as long as we DO change the mols of gas:

#color(blue)(DeltaH) = DeltaU + Delta(nRT)#

#= DeltaU + n_2RT_2 - n_1RT_1#

#= "30.0 L"cdot"atm" + ("0.995 mols" cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot "245 K" - "0.770 mols" cdot "0.082057 L"cdot"atm/mol"cdot"K" cdot "95 K")#

#= color(blue)("44.0 L"cdot"atm")#

By the way, note that

#Delta(PV) ne PDeltaV + VDeltaP#

Actually,

#Delta(PV) = PDeltaV + VDeltaP + DeltaPDeltaV#

In this case the #DeltaPDeltaV# accounts for #10%# of the #DeltaH# value.

Owen Ambrose · Answer

undefined To find the change in enthalpy (ΔH), you can use the equation ΔH = ΔU + Δ(PV). Substituting the given values, you get: ΔH = 30.0 L atm + (4.0 atm - 2.0 atm) * 5.0 L. Solving this equation gives you ΔH = 42.0 L atm. Therefore, the correct answer is (B) 42.

Serenity Allen · Answer

undefined To find the change in enthalpy ($ \Delta H $) for the process, we can use the formula:

$$ \Delta H = \Delta U + \Delta (PV) $$

Where $ \Delta U $ is the change in internal energy and $ \Delta (PV) $ is the work done by the gas. Since the gas is ideal, we can use the formula for work done by an ideal gas:

$$ \Delta (PV) = nR \Delta T $$

Where $ n $ is the number of moles of gas, $ R $ is the ideal gas constant, and $ \Delta T $ is the change in temperature.

First, we need to find the number of moles of gas. We can use the ideal gas law:

$$ PV = nRT $$

For the initial state:

$$ n_1 = \frac{PV}{RT} = \frac{(2.0 \text{ atm})(3.0 \text{ L})}{0.0821 \text{ L atm/mol K} \times 95 \text{ K}} $$

For the final state:

$$ n_2 = \frac{(4.0 \text{ atm})(5.0 \text{ L})}{0.0821 \text{ L atm/mol K} \times 245 \text{ K}} $$

Then, the change in moles is $ \Delta n = n_2 - n_1 $.

Now, we can calculate $ \Delta (PV) $ and then $ \Delta H $:

$$ \Delta (PV) = (\Delta n)R\Delta T $$

$$ \Delta H = \Delta U + \Delta (PV) $$

Calculate $ \Delta H $ using the given values and the calculated values for $ \Delta n $ and $ \Delta (PV) $. The answer should be one of the options provided.

Naomi Aronson · Answer

undefined The change in enthalpy (ΔH) of the process can be calculated using the equation:

ΔH = ΔU + ΔnRT

Where:
ΔU = Change in internal energy
Δn = Change in moles of gas (final moles - initial moles)
R = Ideal gas constant
T = Temperature in Kelvin

Given:
ΔU = 30.0 L atm
Δn = (moles at final state) - (moles at initial state) = (4 atm × 5 L / (0.0821 L atm/K mol × 245 K)) - (2 atm × 3 L / (0.0821 L atm/K mol × 95 K))
R = 0.0821 L atm/K mol

Calculate Δn and substitute the values into the equation to find ΔH.

An ideal gas undergoes a change of state (2.0 atm. 3.0 L, 95 K ) to (4.0 atm. 5.0 L, 245 K) with a change in internal energy, #DeltaU# = 30.0 L atm. The change in enthalpy (#DeltaH#) of the process in L atm is (A) 44 (B) 42.3 (C) ?

Why don't we use the formula for work as nr(change in temp.) to get the work done by taking n=1. When to use (change in pv) and when to use(change in temp.)

Why don't we use the formula for work as nr(change in temp.) to get the work done by taking n=1.
When to use (change in pv) and when to use(change in temp.)