Question

If 6.00 mol argon in a 100 Lvessel initially at 300 K is compressed adiabatically (q = 0) and irreversibly until a final temperature of 450 K is reached, calculate the energy change of the gas, the heat added to the gas, and the work done on the gas.

Short answer

It is found that$∆U=11.2kJ$.

For the adiabatic process, there is no transfer of heat into or out of the system. So,

$\text{q = 0}$

$w=11.2\text{kJ}$.

Step-by-step solution

Given data

The initial temperature of argon gas is$T_1=300\text{K}$.

The final temperature of argon gas is $T_2=450\text{K}$.

Concept of adiabatic process

In thermodynamics, an adiabatic process is one that happens without the transmission of heat and mass between of thermodynamic system and its surroundings.

Calculation of molar heat capacity

Molar heat capacity of a monoatomic gas can be represented with the aid of the formula.

$c_{\text{V}}=\frac{3}{2}R$ …… (1)

Where, R is the universal gas constant and$c_{\text{v}}$is molar heat capacity.

Now, substitute the value of universal gas constant to find the molar heat capacity in equation (1).

$$\begin{gathered} c_v=\frac{3}{2}\times 8.315{\text{JK}}^{-1}{\text{mol}}^{-1} \\ =12.47{\text{JK}}^{-1}{\text{mol}}^{-1} \end{gathered}$$

Calculation of internal pressure

Internal energy change of a gas will be determined by the formula.

$\Delta U=n{c}_{\text{v}}\Delta T$ .…… (2)

Where,$\Delta U$is the change of internal energy of a gas and$\Delta T$is a change in temperature.

The value of$\Delta T$can be determined as follows:

$\Delta T=T_2-T_1$ ……. (3)

Put the value of$T_1$and$T_2$in equation (3).

$$\begin{gathered} \Delta T=450\text{K}-300\text{K} \\ =150\text{K} \end{gathered}$$

Now, substitute the value of given data in equation (2).

$$\begin{gathered} \Delta U=6.00\text{mol}\times 12.47{\text{JK}}^{-1}{\text{mol}}^{-1}\times 150\text{K} \\ =11.2\times {10}^3\text{J} \end{gathered}$$

Calculation of work done

Work done in adiabatic condition can be determined with the help of first law of thermodynamics relation,

$\Delta U=q+w$.

It is known that, for an adiabatic process q = 0 .

Substitute the value of q = 0 and$\Delta U=11.2\text{kJ}$.

$$\begin{gathered} w=\Delta U-q \\ =11.2\text{kJ}-0 \\ =11.2\text{kJ} \end{gathered}$$

Therefore, work done on the gas,$w=11.2\text{kJ}$ .