91Ó°ÊÓ

1. Number of moles,$n=1.80\text{″¾´Ç±ô}$
2. Initial volume of ideal gas,$V_i=3.00{\text{m}}^3$
3. Final volume of ideal gas,$V_f=1.5{\text{″¾}}^3$
4. Temperature,$T=30°\text{C}=30+273=303\text{‿é}$
5. $R=8.31\text{ }\frac{\text{J}}{\text{mol}â‹\dots \text{K}}$
   

Consider the work done of an ideal gas can be explained in two terms. First, if the gas is allowed to expand from the initial volume$\mathbf{}{\mathbf{V}}_i$to final volume$\mathbf{}{\mathbf{V}}_i$while keeping temperature Tof the gas constant, it is called isothermal expansion, and reverse process is known as isothermal compression. In this problem, an ideal gas is compressed at temperature.$\mathbf{}\mathbf{30}\mathbf{°}\mathbf{C}$ For compression,${\mathbf{V}}_f$is less than.${\mathbf{V}}_f$So, the ratio of the volumes is less than unity. Hence, the work done is negative.

Formula is as follow:

$W=nRT\ln \frac{V_f}{V_i}$

Here, W is work done, V is volume, T is temperature, R is universal gas constant and n is number of moles.

To find the amount of energy transferred as heat during isothermal compression process, i.e.$W=Q,$,

$W=nRT\ln \frac{V_f}{V_i}$

$$\begin{gathered} W=\left(1.80\right)\left(8.31\right)\left(303\right)\ln \frac{\left(1.5\right)}{\left(3\right)} \\ W=\left(1.80\right)\left(8.31\right)\left(303\right)(−0.6931) \end{gathered}$$

$W=−3141.53\text{ J}$

Since:$W=Q,$

$$\begin{gathered} \left|Q\right|=|−3141.53| \\ \begin{array}{c}Q=3.1415×{10}^3\text{ J}\\ ≈3.14×{10}^3\text{ J}\end{array} \end{gathered}$$

Hence, the amount of energy transferred as heat during the compressionis$3.14×{10}^3\text{J}$

The negative sign of the work done implies that the heat is transferred from the gas.

Hence,the heat is transferred from the gas.