## Work done by hydrogen

### Task number: 1285

Hydrogen is diatomic gas, whose molar heat capacity at constant volume is

$C_V =\frac{5}{2}R.$

It fills a volume of 100 cm3 at pressure of 51 kPa. Determine the work done by the gas when it expands its volume five times

a) isothermally,

Note.: Consider hydrogen being an ideal gas.

• #### Notation

 V1 = 100 cm3 = 100·10−6 m3 initial hydrogen volume p1 = 51 kPa = 51·103 Pa initial hydrogen pressure V2 = 5V1 volume after expansion $$C_V =\frac{5}{2}R$$ hydrogen molar heat capacity at constant volume Wi = ? work done during the isothermal expansion Wa = ? work done during adiabatic expansion
• #### Hint

Think about how to calculate the work done by hydrogen when the pressure is function of the volume.

• #### Hint a) – Pressure Expression

To express the pressure p as a function of the volume V in the case of an isothermal process, the so-called Boyle-Mariotte law has to be applied.

• #### Analysis a)

The integral calculus must be used so that we can calculate the work of the hydrogen, because the pressure is a function of the volume.

The hydrogen pressure is expressed from the Boyle-Mariotte law, which is valid for isothermal processes. The obtained feature is integrated by the volume and the initial and the final hydrogen volumes are used as limits of the integration.

• #### Solution a)

At constant pressure, the performed work W is given by W = p(V2 − V1).

However, the pressure is not constant in our case, which means that more general relation is required

$W_i = \int\limits_{V_1}^{V_2}p\, \text{d}V,$

where V1 is the initial hydrogen volume and V2 is its volume after the expansion.

Therefore, the pressure p has to be formulated as a function of the volume V. An obvious consequence of the isothermal process state equation – the Boyle‑Mariotte law, can help with it. It states:

$p_1V_1 = pV.$

It is possible to express the pressure p immediately:

$p = \frac{p_1V_1}{V}.$

The next step is the actual integration

$W_i = \int\limits_{V_1}^{V_2}p\, \text{d}V = \int\limits_{V_1}^{5V_1}\frac{p_1V_1}{V}\, \text{d}V =$

Constants are factored out

$=p_1V_1 \int\limits_{V_1}^{5V_1}\frac{1}{V}\, \text{d}V =$

After integration and recall of the limits

$=p_1V_1[\ln\,V]_{V_1}^{5V_1} = p_1V_1\,\ln \frac{5V_1}{V_1}= p_1V_1\,\ln 5.$
• #### Hint b) – Pressure Expression

To express the pressure p as a function of the volume V in the case of an adiabatic process, the Poisson’s law has to be applied.

• #### Hint b) – Poisson

To calculate the Poisson’s ratio κ the following relation could be applied

$\kappa = \frac{C_p}{C_V},$

where Cp is the molar heat capacity at constant pressure and CV is the molar heat capacity at constant volume. The so-called Mayer’s relation is held between the molar heat capacities.

$C_p=C_V+R,$

where R is the molar gas constant.

• #### Analysis b)

As in the previous task section, the integral calculus must be used here so that we can find the work done by the hydrogen, because the pressure is a function of the volume.

This time, the hydrogen pressure is expressed by the Poisson’s law which is valid for adiabatic processes. The received function is integrated by the volume. The initial and the final volume are used as integral limits.

Finally, it is necessary to evaluate the Poisson’s constant from the Mayer’s relation and the relation between the Poisson’s constant and the molar heat capacity at the constant pressure and the constant volume.

• #### Solution b)

As a consequence of a variable pressure, the following equation is applied again to calculate the work Wa:

$W_a = \int\limits_{V_1}^{V_2}p \, \text{d}V,$

where V1 is the initial hydrogen volume and V2 is the volume after the expansion.

The Poisson’s law for adiabatic (heat isolated) processes expresses the pressure p as a function of the volume V:

$p_1V_{1}^{\kappa} = pV^{\kappa}.$

It is possible to derive the pressure p immediately:

$p = \frac{p_1V_1^{\kappa}}{V^{\kappa}}.$

By combining the received formulation with the formula for work

$W_a = \int\limits_{V_1}^{V_2}p \, \text{d}V = \int\limits_{V_1}^{5V_1}\frac{p_1V_1^{\kappa}}{V^{\kappa}} \, \text{d}V =$

The constants are factored out of the integral

$= p_1V_1^{\kappa} \int\limits_{V_1}^{5V_1}\frac{1}{V^{\kappa}} \, \text{d}V =$

The integral is solved and the limits are recalled

$= p_1V_1^{\kappa}\frac{1}{-\kappa + 1}\left[V^{-\kappa + 1}\right]_{V_1}^{5V_1}= \frac{p_1V_1^{\kappa}}{-\kappa + 1}\left[(5V_1)^{-\kappa + 1} - V_1^{-\kappa + 1}\right].$

After simplification, we get

$W_a = p_1V_1\,\frac{5^{\kappa-1}-1}{\kappa-1}.$

However, the Poisson’s constant κ of the given gas is still unknown. To find it, its definitional formula is needed

$\kappa = \frac{C_p}{C_V},$

where Cp means the molar heat capacity at constant pressure and CV the molar heat capacity at constant volume, as well as the Mayer’s law

$C_p = C_V + R,$

where R denotes the molar gas constant.

After substitution

$\kappa = \frac{C_V + R}{C_V} = \frac{\frac{5}{2}R + R}{\frac{5}{2}R} = \frac{7}{5}.$

The received value of Poisson’s constant κ is applied to the expression of the work

$W_a = p_1V_1\,\frac{5\cdot(5^{2/5}-1)}{2}.$
• #### Numerical Evaluation

a) The work for isothermal expansion

$W_i = p_1V_1 \ln\,5 = 51\cdot{10^3}\cdot 100\cdot{10^{-6}}\cdot \ln\,5\, \,\mathrm{J} \dot{=} 8.21\, \mathrm{J}$

b) The work for adiabatic expansion

$W_a = p_1V_1\,\frac{5\cdot(5^{2/5}-1)}{2} = 51\cdot{10^3}\cdot 100\cdot{10^{-6}}\cdot \frac{5\cdot(5^{2/5}-1)}{2}\, \mathrm{J}$ $W_a \dot{=} 6.05\, \mathrm{J}$