Question

if 2 moles of ideal moatomic gas at temp To is mixed with 4 moles of another ideal moatomic gas at temp 2To,then the temp of mixture is

Answer 1

Answer to Question #72392, Physics / Mechanics | Relativity

Given,

$$T _ {1} = T _ {0} \quad n _ {1} = 2$$ $$T _ {2} = 2 T _ {0} \quad n _ {2} = 4$$

Let the volume of the gases are $V$ and mixed at a container with same volume $V$ .

I have assumed the volume because without knowing the information about volume this problem can't be solved.

So, $V_{1} = V_{2} = V$

Now, their pressure before mixing are -

$$P _ {1} = \frac {n _ {1} R T _ {1}}{V _ {1}}$$ $$P _ {2} = \frac {n _ {2} R T _ {2}}{V _ {2}}$$

After mixing,

$$n = \quad n _ {1} + \quad n _ {2}$$

Now, from the Daltons low of partial pressure

Net pressure $P = P_{1} + P_{2}$

$$= \frac {n _ {1} R T _ {1}}{V _ {1}} + \frac {n _ {2} R T _ {2}}{V _ {2}}$$

Let after mixing the temperature become T

Answer to Question #72392, Physics / Mechanics | Relativity

Now,

$$\begin{array}{l} T = \frac {P V}{n R} \\ = - \frac {\left(\frac {n _ {1} R T _ {1}}{V _ {1}} + \frac {n _ {2} R T _ {2}}{V _ {2}}\right) V}{n R} \\ = - \frac {\left(\frac {2 R T _ {0}}{V} + \frac {4 R 2 T _ {0}}{V}\right) V}{6 R} \\ \end{array}$$

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