Question

Consider 111 J of heat added to 30.3 g of Ne on STP. Determine w, ΔE, and ΔH if the heat is transferred at constant volume. Do the same determinations if the heat is transferred at constant pressure. Calculate the final temperature under each condition.

Short answer

The value of w, ΔE, and ΔHfor the first case (constant volume) will be 0, 111J and 185J respectively.The value of w, ΔE, and ΔHfor the second case (constant pressure) will be -44.4 J, 66.6 J and 111J respectively.

The final temperature for the first case (constant volume) will be 278.94 K. The final temperature for the second case (constant pressure) will be 276.6 K.

Step-by-step solution

Introduction

Ne is a monoatomic gas.

For monoatomic gas we can write

$$\begin{gathered} C_v=\frac{3}{2}R \\ {C}_p=\frac{5}{2}R \end{gathered}$$

Let the work done by the system be denoted by the symbol w. The pressure and volume are denoted by P and V respectively. The amount of heat transferred is q. The internal energy be ΔE and enthalpy be ΔH.

No of moles of Ne present will be

$\begin{array}{rcl}n& =& 30.3g\times \frac{1mol}{20.18g}\\ & =& 1.5mol\end{array}$

As it is stated that the gas is at STP therefore, it can be stated that pressure P=1 atm and temperature T=273 K. So, it was assumed that the initial temperature T₁=273K and the final temperature will be T₂

At constant volume

$$\begin{gathered} ∆E=q_v \\ =n{C}_v∆T \\ =n\frac{3}{2}R∆T \end{gathered}$$

At constant pressure

$$\begin{gathered} ∆H=q_p \\ =n{C}_p∆T \\ =n\frac{5}{2}R∆T \end{gathered}$$

Therefore,

role="math" localid="1660048803703" $$\begin{gathered} \frac{∆H}{∆E}=\frac{nR{\displaystyle \frac{5}{2}}∆T}{nR\frac{3}{2}∆T} \\ \frac{∆H}{∆E}=\frac{5}{3} \\ ∆H=\frac{5}{3}∆E \end{gathered}$$

Determine the values at constant volume

For the first case we need to determine w, ΔE and ΔH if heat is transferred at constant volume. It has been stated that 111J of heat is added to the system.

Given

q=111 J

At constant volume

$\begin{array}{rcl}q& =& n{C}_v∆T\\ 111J& =& n\left(\frac{3}{2}R\right)\left(T_2-T_1\right)\\ 111J& =& 1.5mol\times \frac{3}{2}\times 8.314Jmo{l}^{-1}{K}^{-1}\times \left(T_2-273\right)K\\ T_2& =& 278.94K\end{array}$

Therefore, the final temperature for the first case will be 278.94 K

At constant volume$∆V=0$

Therefore,

$\begin{array}{rcl}w& =& P∆V\\ & =& 0\end{array}$

$\begin{array}{rcl}∆E& =& q+w\\ & =& 111J+0\\ & =& 111J\end{array}$

$\begin{array}{rcl}∆H& =& \frac{5}{3}∆E\\ & =& \frac{5}{3}\times 111J\\ & =& 185J\end{array}$

Determine the values at constant pressure. 

For the second case we need to determine w, ΔE and ΔH if heat is transferred at constant pressure. It has been stated that 111J of heat is added to the system.

Given

q=111 J

At constant pressure the following can be obtained

$\begin{array}{rcl}∆H& =& q\\ q& =& n{C}_p∆T\\ 111J& =& n\left(\frac{5}{2}R\right)\left(T_2-T_1\right)\\ 111J& =& 1.5mol\times \frac{5}{2}\times 8.314Jmo{l}^{-1}{K}^{-1}\times \left(T_2-273\right)K\\ T_2& =& 276.6K\end{array}$

Therefore, the final temperature for the second case will be 276.6 K

$\begin{array}{rcl}∆E& =& \frac{3}{5}∆H\\ & =& \frac{3}{5}\times 111J\\ & =& 66.6J\end{array}$

$\begin{array}{rcl}∆E& =& q+w\\ 66.6J& =& 111J+w\\ w& =& -44.4J\end{array}$