Question

Suppose 2.00 mol of an ideal, monatomic gas is initially at a pressure of 3.00 atm and a temperature T=350 K. It is expanded irreversibly and adiabatically (q=0) against a constant external pressure of 1.00 atm until the volume

has doubled.

(a) Calculate the final volume.

(b) Calculate w, q, and$\mathbf{\Delta U}$
for this process, in joules.

(c) Calculate the final temperature of the gas.

Short answer

(a) The final volume is 38.2 L.

(b)The value of w is 1935 J, q is equal to 0 and the value of $∆U=-1935J$.

(c) The final temperature of gas is 272.4 K.

Step-by-step solution

Ideal gas law:

The ideal gas law is also termed general gas equation. The behavior of a particular gas can be studied using the ideal gas equation. The equation relating ideal gas is:$\mathbf{PV}\mathbf{=}\mathbf{nRT}$

Calculating the final volume, work done, heat energy, internal energy and final temperature

(a) According to the problem, the given gas is ideal and monoatomic.

We know that, PV=nRT

Where p is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant and T is the temperature

Substituting the above values in the equation to find the volume,

$$\begin{gathered} \text{V =}\frac{\text{nRT}}{\text{P}} \\ \text{=}\frac{{\text{2×0.082 L atm mol}}^{\text{- 1}}{\text{K}}^{\text{- 1}}\text{×350 K}}{\text{3 atm}} \\ \text{= 19.1L} \end{gathered}$$

$$\begin{gathered} \text{Final Volume = 2×19.1 L} \\ \text{= 38.2 L} \end{gathered}$$

(b)The work done by the system can be found as:

$$\begin{gathered} \text{W = - P}\left({\text{V}}_{\text{2}}{\text{-V}}_{\text{1}}\right) \\ \text{= - 1 atm}\left(\text{38.2 - 19.1}\right)\text{L} \\ \text{= - 19.1 L atm} \end{gathered}$$

The work in atom can be converted to joules as:

$$\begin{gathered} \text{W = - 19.1 L atm}\frac{\text{101.325 J}}{\text{L atm}} \\ \text{= - 1935 J} \end{gathered}$$

As the process is adiabatic, the total heat involved, q=0.

From the first law of thermodynamics,

$$\begin{gathered} \text{dU = dq + dW} \\ \text{dU = 0 - 1935J} \\ \text{= - 1935J} \end{gathered}$$

(c) As the gas is monoatomic, so the degrees of freedom of the gas is 3. Hence, the total molar internal energy of the gas, ${\text{U}}_{\text{m}}\text{=}\frac{\text{3RT}}{\text{2}}$

Where, R is the universal gas constant, $\text{8.314 J mol}{}^{\text{- 1}}{\text{K}}^{\text{- 1}}$ and T is the Kelvin temperature.

The molar heat capacity of the gas at constant volume,

$$\begin{gathered} {\text{C}}_{\text{vm}}\text{=}\frac{\text{3R}}{\text{2}} \\ {\text{= 12.47JK}}^{\text{- 1}} \end{gathered}$$

For adiabatic process,

$$\begin{gathered} {\text{U = nC}}_{\text{vm}}\text{×}\left({\text{T}}_{\text{2}}{\text{-T}}_{\text{1}}\right) \\ {\text{- 1935 = 2×12.47JK}}^{\text{- 1}}\left({\text{T}}_{\text{2}}{\text{-T}}_{\text{1}}\right) \\ \left({\text{T}}_{\text{2}}{\text{-T}}_{\text{1}}\right)\text{=}\frac{\text{- 1935J}}{{\text{2×12.47JK}}^{\text{- 1}}} \\ \text{= - 77.6K} \end{gathered}$$

The value of ${\text{T}}_{\text{1}}\text{= 350K}$.

The value of${\text{T}}_{\text{2}}$ can be found as:

$$\begin{gathered} {\text{T}}_{\text{2}}\text{- 350K = - 77.6K} \\ {\text{T}}_{\text{2}}\text{= - 77.6K + 350K} \\ \text{= 272.4K} \end{gathered}$$

The final temperature of gas is 272.4 K.