A 20.0 -\,L volume of an ideal gas in a cylinder with a piston is at a pressure of 3.2 atm. Enough weight is suddenly removed from the piston to lower the external pressure to 1.6 atm. The gas then expands at constant temperature until its pressure is 1.6 atm.

1. Find Delta E for this change in state in J
2. Find Delta H for this change in state in J
3. Find q for this change in state in J
4. Find w for this change in state in J

The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.

The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.

P₁ × V₁ = P₂ × V₂

3.2 atm × 20.0 L = 1.6 atm × V₂

V₂ = 40 L

The work (w) can be calculated using the following expression.

w = - P . ΔV

where,

P is the external pressure for which the process happened

ΔV is the change in the volume

w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J

The change in the internal energy is:

ΔE = q + w

0 = q + w

q = - w = 3.2 × 10³ J

aliasger2709 aliasger2709 · Answer 2 · 2022-03-16T08:09:14+01:00

An ideal gas is a hypothetical gas that follows Charles, Boyle's, Gay-Lussac's, and Avogadro's law. [tex]\rm \Delta E[/tex] and [tex]\rm \Delta H[/tex] is 0 J, q is [tex]3.2 \times 10^{3} \;\rm J[/tex] and w is [tex]-3.2 \times 10^{3} \;\rm J.[/tex]

What is an ideal gas equation?

An ideal gas equation is the theoretical gas equation that gives the relation of the entropy, enthalpy, temperature, volume, moles and pressures of the gas.

When the temperature of the system is zero or constant then the change in the enthalpy will be equal to zero hence, change in the internal energy [tex]\rm \Delta E[/tex] and change in the enthalpy [tex]\rm \Delta H[/tex] = 0.

According to Boyle's law final volume can be calculated as:

[tex]\begin{aligned} \rm P_{1} \times V_{1} &= \rm P_{2} \times V_{2}\\\\3.2 \;\rm atm \times 20.0 \;\rm L &= 1.6 \;\rm atm \times V_{2}\\\\&= 40\;\rm L\end{aligned}[/tex]

Work (w) can be calculated as:

[tex]\begin{aligned}\rm Work (w) &=\rm -P\times \Delta V\\\\&= -1.6 \;\rm atm \times (40L - 20.0L)\\\\&= -3.2 \times 10^{3} \;\rm J\end{aligned}[/tex]

Change in internal energy (q) can be calculated as:

[tex]\begin {aligned} \rm \Delta E &= \rm q + w\\\\0& = \rm q + w\\\\\rm q &= \rm - w = 3.2 \times 10^{3}\;\rm J\end{aligned}[/tex]

Therefore, change in the internal energy and the change in the enthalpy is 0, work is [tex]-3.2 \times 10^{3} \;\rm J[/tex] and q is [tex]3.2 \times 10^{3} \;\rm J.[/tex]

Learn more about internal energy and enthalpy here:

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