Question

Calculate$\mathit{q}\mathbf{,}\mathit{w}\mathbf{,}\mathbf{∆}\mathit{E}\mathbf{,}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathbf{∆}\mathit{H}$ for the process in which 88.0 g of nitrous oxide gas$\left(N_{2}O\right)$ is cooled from $165°Cto55°C$at a constant pressure of 5.00 atm. (Hint ;${\mathit{C}}_{\mathbf{p}}$ for $N_{2}O=38.7J/Kmol$)

Short answer

$$\begin{gathered} q=8.51kJ \\ w=1.829kJ \\ ∆E=10.339kJ \\ ∆H=8.51kJ \end{gathered}$$

Step-by-step solution

Calculate moles of N2O

In this we will find the moles in given mass of compound and then moles in 88g .

Given, Mass of= 88.0 g

$$\begin{gathered} Molesingivenmassofacompound=\frac{Mass}{Molarmass} \\ Molesin88gof{N}_{2}O=\frac{88g}{44g/mol}=2moles \end{gathered}$$

Calculation for q

In this step we will find the heat at a constant pressure .

Here, n = 2 moles

$C_p$= 38.7 J/ K mol

$∆T=(165-55)°C=110°C$

At constant pressure

$$\begin{gathered} q=n{C}_p∆T \\ =2mol\times 38.7J/K-mol\times 100K \\ =8,514J=8.51kJ \end{gathered}$$

Calculation for w

In this we will find the amount of work done using pressure and change in volume.

$$\begin{gathered} w=P∆V=nR∆T=2\times 8.3145\times 100 \\ =1829.08J=1.829kJ \end{gathered}$$

Where,

$n=2moles,R=8.314J/Kmol,∆T=110K$

So,

Calculation of  ∆E

$∆E=q+w=8.51+1.829kJ=10.339kJ$

Calculation of ∆H

$∆H=n{C}_p∆T=q=8.51kJ$