Question

Consider a rigid, insulated box containing 0.400 mole of he(g) at 20.08c and 1.00 atm in one compartment and 0.600 mole of n2(g) at 100.08c and 2.00 atm in the other compartment. these compartments are connected by a partition that transmits heat. what is the final tempera-ture in the box at thermal equilibrium? [for he(g), cv 5 12.5 j k 21mol 21 ; for n2(g), cv 5 20.7 j k 21mol 21

Answer 1

Answer:

The value of final temperature of the mixture of gases when the partition is removed $T_{3}$ = 341.08 K

Explanation:

For box 1

Mass $m_{1}$ = 0.4 mole

Temperature $T_{1}$ = 20.08 °c = 293.08 K

Pressure $P_{1}$ = 1 atm = 101.325 K pa

For box 2

Mass $m_{2}$ = 0.6 mole

Temperature $T_{2}$ = 373.08 K

Pressure $P_{2}$ = 2 atm = 202 K pa

When the partition removed, both gases mixes together. then the final mass

$m_{3}$ = $m_{1}$ + $m_{2}$

$m_{3}$ = 0.4 + 0.6

$m_{3}$ = 1.0 mole

This is the final mass in the container.

Final temperature is given by the equation,

$m_{1}$ $T_{1}$ + $m_{2}$ $T_{2}$ = $m_{3}$ $T_{3}$

Put all the values in the above equation we get,

0.4 × 293.08 + 0.6 × 373.08 = 1  × $T_{3}$

$T_{3}$ = 341.08 K

This is the value of final temperature of the mixture of gases when the partition is removed.