First slide

Questions on calorimetry without phase change

Question

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T_o, while box B contains one mole of helium at temperature (7/3) T₀.The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (Ignore the heat capacity of boxes). Then the final temperature of the gases, T_f in terms of T_o is

Easy

A

$T_{f} = \frac{3}{7} T_{0}$
B

$T_{f} = \frac{7}{3} T_{0}$
C

$T_{f} = \frac{3}{2} T_{0}$
D

$T_{f} = \frac{5}{2} T_{0}$

Solution

Here, the change in internal energy of the system is zero. The increase in the internal energy of one box will be equal to the decrease in internal energy of the other box. The boxes are shown in fig.
The change in internal energy is given by
$\begin{aligned} {dU}_{A} = 1 \times \frac{5 R}{2} (T_{f} - T_{0}) \\ {dU}_{B} = 1 \times \frac{3 R}{2} (T_{f} - \frac{7}{3} T_{0}) \end{aligned}$
Now, ${dU}_{A} + {dU}_{B} = 0$
$∴ \frac{5 R}{2} (T_{f} - T_{0}) + \frac{3 R}{2} (T_{f} - \frac{7}{3} T_{0}) = 0$
or $5 T_{f} - 5 T_{0} + 3 T_{f} - 7 T_{0} = 0$
or $8 T_{f} = 12 T_{0} or T_{f} = \frac{12}{8} T_{0}$
$∴ T_{f} = \frac{3}{2} T_{0}$

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