91影视

Amount of diatomic gas$=0.020mol$

Temperature$={20}^{鈭�}C$

Diatomic gas compressed from$1500c{m}^{3}to500c{m}^3$

As we can see from the first law of thermodynamics, we can figure out how much heat a gas adds

$螖E=Q+W鈬�Q=螖E-W$

As a consequence, it is necessary to calculate the change in internal energy as well as the work,

Change in internal energy is given as,

$螖E=n{C}_v螖T$

It is therefore needed to find the change in temperature. Since no parameter among $(p,V,T)$remain constant, we can only say

$\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}鈬�T_2=\frac{p_2{V}_2}{p_1{V}_1}{T}_1$

The pressure being proportional to the inverse of the volume squared since pressure times volume to the second power remains constant, we can write:

$p{V}^2=\text{const.}鈬�p鈭�\frac{1}{V^2}$

Hence the pressure ratio as,

$\frac{p_2}{p_1}=\frac{V_1^2}{V_2^2}$

Substituting, we have

$T_2=\frac{V_1^2}{V_2^2}\frac{V_2}{V_1}{T}_1=\frac{V_1}{V_2}{T}_1$

The temperature difference as

$螖T=T_2-T_1=\left(\frac{V_1}{V_2}-1\right)T_1=\frac{V_1-V_2}{V_2}{T}_1$

Change in internal energy as,

$螖E=n{C}_v螖T=\frac{V_1-V_2}{V_2}n{C}_v{T}_1$

An evaluation of work in any process is based on dividing the integral of the pressure across the volume by the integral of the pressure.

$W=-{鈭�}_{V_1}^{V_2}pdV$

Considering what we've been told and what we've already written down, we have this information,

$p{V}^2=\text{const.}=K$

$K$be constant,

$p=\frac{K}{V^2}$

Work will become,

$W=-{鈭�}_{V_1}^{V_2}\frac{K}{V^2}dV=-K{鈭�}_{V_1}^{V_2}{V}^{-2}dV$

By integrating,

$W=-K{\left[\frac{V^{-1}}{-1}\right]}_{V_1}^{V_2}=K{\left[\frac{1}{V}\right]}_{V_1}^{V_2}=K\left(\frac{1}{V_2}-\frac{1}{V_1}\right)=K\frac{V_1-V_2}{V_1{V}_2}$

Using the equation for the process applied in the initial state, we can determine each parameter in the expression for work after expressing $K,$

$p_1{V}_1^2=K$

Our knowledge of the initial pressure is more limited.

However, since we know the initial pressure, we can write it out according to the ideal gas law:

$p_1{V}_1=nR{T}_1鈬�p_1=\frac{nR{T}_1}{V_1}$

Substituting the result,

role="math" localid="1648203934671" $K=\frac{nR{T}_1}{V_1}{V}_1^2=nR{T}_1{V}_1$

We obtain the following result by substituting our result in the expression for the work:

$W=nR{T}_1\frac{V_1-V_2}{V_2}$

After having expressed both the change in internal energy and the work, in line with the first thing we wrote in this exercise, we can now translate that into an expression for the heat:

$Q=螖E-W=n{C}_v{T}_1\frac{V_1-V_2}{V_2}-nR{T}_1\frac{V_1-V_2}{V_2}$

By Factorization,

$Q=n{T}_1\frac{V_1-V_2}{V_2}\left(C_v-R\right)$

By Substituting numerically,

$Q=0.02路293路\frac{(1.5-0.5)}{0.5}(20.8-8.31)=150\mathrm{J}$

Hence, the amount of heat energy is added during this process is$150J.$