Question

Hydrogen behaves as an ideal gas at temperatures greater than 200 K and at pressures less than 50 atm. Suppose 6.00 mol hydrogen is initially contained in a 100-L vessel at a pressure of 2.00 atm. The average molar heat capacity of hydrogen at constant pressure, ${\mathbf{c}}_{\mathbf{P}}$, is$\mathbf{29}\mathbf{.}\mathbf{3}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$ in the temperature range of this problem. The gas is cooled reversibly at constant pressure from its initial state to a volume of 50.0 L. Calculate the following quantities for this process.

(a) Temperature of the gas in the final state,${\mathbf{T}}_{\mathbf{2}}$

(b) Work done on the gas, w,in joules

(d) Heat absorbed by the gas, q, in joules

Short answer

(a) The temperature of the gas in the final state is 203.1 K.

Step-by-step solution

Given Information.

The temperature is greater than 200 K.

The pressure is less than 50 atm.

The number of moles of hydrogen is 6.00 mol.

The volume of the vessel is 100 L.

The pressure is 2.00 atm.

The average molar heat capacity is${\text{29.3JK}}^{\text{- 1}}{\text{mol}}^{\text{- 1}}$.

The final volume is 50.0 L.

Temperature of the gas.

The temperature of the gas in the final state can be obtained from the formula:

$\frac{{\mathbf{\text{V}}}_{\mathbf{\text{1}}}}{{\mathbf{\text{V}}}_{\mathbf{\text{2}}}}\mathbf{\text{=}}\frac{{\mathbf{\text{T}}}_{\mathbf{\text{1}}}}{{\mathbf{\text{T}}}_{\mathbf{\text{2}}}}$

${\mathbf{\text{T}}}_{\mathbf{\text{2}}}{\mathbf{\text{=T}}}_{\mathbf{\text{1}}}\mathbf{\text{×}}\frac{{\mathbf{\text{V}}}_{\mathbf{\text{2}}}}{{\mathbf{\text{V}}}_{\mathbf{\text{1}}}}$

Here,

${\mathbf{\text{T}}}_{\mathbf{\text{1}}}$and${\mathbf{\text{T}}}_{\mathbf{2}}$ denote the initial and final temperatures.

${\mathit{V}}_{\mathbf{\text{1}}}$and${\mathit{V}}_{\mathbf{2}}$ denote the initial and final volumes.

Computing the temperature of the gas in the final state.Subpart (a)

The ideal gas equation is:

$\text{PV = nRT}$

The initial value of temperature can be found as;

$$\begin{gathered} {\text{T}}_{\text{1}}\text{=}\frac{{\text{PV}}_{\text{1}}}{\text{n R}} \\ \text{=}\frac{\text{2 atm×100 L}}{{\text{6.00 mol×0.08206 L atmK}}^{\text{- 1}}{\text{mol}}^{\text{- 1}}} \\ \text{= 406.2 K} \end{gathered}$$

At constant pressure, P, n, and R remain unchanged.

As;

$\frac{{\text{V}}_{\text{1}}}{{\text{V}}_{\text{2}}}\text{=}\frac{{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}}$

Therefore,

$$\begin{gathered} {\text{T}}_{\text{2}}{\text{=T}}_{\text{1}}\frac{{\text{V}}_{\text{2}}}{{\text{V}}_{\text{1}}} \\ \text{= 406.2 K}\frac{\text{50.0 L}}{\text{100 L}} \\ \text{= 203.1 K} \end{gathered}$$

Hence, the temperature of the gas in the final state is 203.1 K.