Answer to Question #106639 in Molecular Physics | Thermodynamics for Erickson Carno

Question

A 32.4 L sealed cylinder contains 120. moles of helium gas at 44.8 atm. What iis its temperature (in Celsius)?

Answer 1

Solution:

When using the van der Waals equation

$$(p+ \frac {a\times v^2}{V^2})(V-b\times v)=v\times R \times T$$

we use the following values, taking into account that helium is in the cylinder

$$p=44.8\times 101325.011=4.54\times 10^6 \frac {N}{m^2}$$

$$a=3.38 \times 10^{-3} \frac {N\times m^4}{mole^2}$$

$$v=120 mole$$

$$V=3.24\times 10^{-2} m^3$$

$$b=2.36 \times 10^{-5} \frac {m^3}{mole}$$

$$R=8.314 \frac {Dg}{mole \times K}$$

$$T= \frac{(p+\frac {av^2}{V^2}) (V-bv)}{vR}$$

$$T=136 K$$

$$t=T-273=-137 C$$

Answer: -137 (in Celsius)