I am trying to find the enthalphy of Helium if its internal energy is 200 kJ/kg. The answer is given as 333.42 kJ/kg, but I keep getting the wrong answer.

I've tried the following:

$\begin {align} H &= U + PV \\ H &= U + mRT \\ \end {align}$

Universal Gas Constant, $R_u = 8.314 \:\mathrm{\dfrac{J}{K \cdot mol}}$

Individual Gas Constant, $R_i = 8.314 \:\mathrm{\dfrac{J}{K \cdot mol}} \div 4 \:\mathrm{\dfrac{g}{mol}} = 2.0785 \:\mathrm{\dfrac{kJ}{kg \cdot K}}$

$\begin {align} H &= 200 \:\mathrm{kJ/kg} + 1\:\mathrm{kg}*2.0785 \:\mathrm{kJ/kgK}*273.15 \:\mathrm{K} \\ H &= 767.74 \;...\\ \end {align}$

Why is my answer different?


First, notice that the problem statement does not specify a temperature.

The internal energy and enthalpy are given by: $$ U = m c_vT $$ $$ H = m c_pT $$


$$ H = \frac{c_p}{c_v}U $$

For a monatomic gas, such as helium, the ratio $c_p/c_v$ is very nearly $5/3$. It follows that:

$$ H = \frac{5}{3} \times 200 = 333.33$$

This is very close to your expected answer. If you were to use the precise values of $c_p$ and $c_v$ given by your textbook, I expect that you will get your textbook's answer of 333.42.

  • $\begingroup$ Where did that $R$ came from in your $U$ and $H$ formulas? $\endgroup$ – Algo Jul 6 '15 at 6:29
  • $\begingroup$ @algo Thanks for spotting that typo. (I had originally written $U=m\frac{3}{2}RT$....). Fixed. $\endgroup$ – John1024 Jul 6 '15 at 6:35

Following your method: You can't assume that temperature of Helium at that particular state is $273.15\ K$, but we can calculate it as follows:

$$u = C_{v}\Delta T$$ $$\Delta T = \frac{u}{C_v} = \frac{200}{3} = 66.66667\ K$$ $$h = C_{p}\Delta T = 5*66.66667 = 333.3 \ kJ/kg$$

As John said, Values from your text book of $C_p$ and $C_v$ should yield a value closer to your expected answer.


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