Question

Difference in final temperatures? Please show all steps

Two containers (call them A and B) with the same initial volume, V 31 L, hold the same amount of monatomic ideal gas, say helium, at the same initial temperature T = 301 K and the same initial pressure Pi 1 atm. Container A holds the gas at a constant volume and Container B holds the gas at a constant pressure If you add Q 5389 J of heat to both containers, what is the difference in the final temperatures after the heat is added, TfA TB? This is the difference in the final temperatures between Container A, a constant Volume process, and Container B, a constant Pressure process, when the heat added is the same Give your answer in Kelvin to three significant digits. The amount of gas in the containers does not change.

Answer 1

DATA:

Solution:

According to the gas ideal equation:

Where:

is the Pressure in the containers, which are equal.

is the volume

is the number of mole

is the ideal gas constant, which has a value of:

$R=0.0821\,\frac{atm.L}{mol.K}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (2)$

is the temperature.

Now, isolating from equation (1) we have:

$n=\frac{PV}{RT}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (3)$

The number of mole in both containers are the same, so:

$n_{A}=n_{B}=\frac{PV}{RT}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (4)$

Inserting (2) and data given values into equation (4) we have:

$n_{A}=n_{B}=\frac{(1\,atm)(31\,L)}{(0.0821\,\frac{atm.L}{mol.K})(301\,K)}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Solving we obtain:

Now, the heat added on each container is defined as:

• For container A . Constant Volume we have:

​ $Q=n_{A}C_{V}\Delta T$

or

$Q=n_{A}C_{V}\left ( T_{f,A}-T_{i,A} \right )$

Isolating $T_{f,A}$ from equation above we have:

$T_{f,A}=T_{i,A}+\frac{Q}{n_{A}C_{V}}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (6)$

Where, $C_{V}$ is the molar specific heat at a constant volume, which has a value of:

$C_{V}=\frac{3}{2}R=12.5\,\frac{J}{mol.K}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (7)$

Now, replacing (5), (7) and data given values into equation (6) we get:

$T_{f,A}=301\,K+\frac{5389\,J}{(1.26\,mol)(12.5\, \frac{J}{mol.K})}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Solving we obtain:

${\color{Red} T_{f,A}=643\,K}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (8)$

• For Container B, Constant Presurre we have:

$C_{P}=\frac{5}{2}R=20.8\,\frac{J}{mol.K}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (9)$

Therefore, the final temperature on the container B is:

$T_{f,B}=T_{i,B}+\frac{Q}{n_{B}C_{P}}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (10)$

Replacing (5) , (9) and data given values into equation (10) we have:

$T_{f,B}=301\,K+\frac{5389\,J}{(1.26\,mol)(20.8\,J)}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (11)$

Solving we obtain:

${\color{Red} T_{f,B}=507\,K}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (12)$

Now, the difference in the final temperatures is :

$T_{f,A}-T_{f,B}=643\,K-507\,K$

Solving we obtain:

${\color{Blue} T_{f,A}-T_{f,B}=136\,K}$