Question

Three moles of an ideal gas
Cv=5cal deg-1 moe-1) at 10 atm
and 0°C are converted to 2 atm
at 50°C. Find delta H and delta U for the
change,​

Answer 1

Explanation:

1) \Delta E=nCvdTΔE=nCvdT

n=3

T2=50oC=323,15K

T1=0oC=273.15K

\Delta E=nCvdT=3\times5\times(323.15-273.15)=750 calΔE=nCvdT=3×5×(323.15−273.15)=750cal

2)

\Delta H=nCpdT=n(Cv+R)dTΔH=nCpdT=n(Cv+R)dT

\Delta H=n(Cv+R)dT=3\times(5+2)\times(323.15-273.15)=1050 calΔH=n(Cv+R)dT=3×(5+2)×(323.15−273.15)=1050cal

Answer 2

Answer:

Explanation:

Concept:

The change in the system's enthalpy during a chemical reaction is equal to the change in the system's internal energy plus the change in the volume of the system times the pressure product.

Therefore, enthalpy and heat are two quite distinct concepts. Not just under continual pressure does it apply. The change in enthalpy of the material, on the other hand, is equal to the quantity of heat added to the material throughout the process if a procedure is performed on a material under constant pressure.

Given:

At $10.0 atm$ and $00$, three moles of an ideal gas $(CV = 5 cal deg^{-1} mol^{-1} )$ are changed to $2.0 atm$ at $500$.

Find:

To find the value of $\triangle H$ and $\triangleU$$\triangle U$ for the change.

Solution:

(1) Find the value of $\triangle H$.

$\Delta E=n C v d T$

$\mathrm{n}=3$

$\mathrm{T}_{2}=50^{\circ} \mathrm{C}=323,15 \mathrm{~K}$

$\mathrm{T}_{1}=0^{\circ} \mathrm{C}=273.15 \mathrm{~K}$

$\Delta E=n C v d T=3 \times 5 \times(323.15-273.15)=750 \mathrm{cal}$

(2) Find the value of $\triangle U$.

$\Delta H=n C p d T=n(C v+R) d T$

$\Delta H=n(C v+R) d T=3 \times(5+2) \times(323.15-273.15)=1050 \mathrm{cal}$.

Therefore, the value of $\triangle H$ and $\triangleU$$\triangle U$ for the change are $750 cal$ and $1050 cal$.

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