Question

A 0.773 mol sample of Xe(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the temperature of the gas by 20.1 K. Assuming ideal gas behavior, calculate the amount of heat in joules (q) required to affect this temperature change and the total change in internal energy, ?U. Note that some books use ?E as the symbol for internal energy instead of ?U.
q=___J ?U=___J

Answer 1

Answer : The value of $q\text{ and }\Delta U$ is 193.8 J and 193.8 J respectively.

Explanation : Given,

Moles of sample = 0.773 mol

Change in temperature = 20.1 K

First we have to calculate the heat absorbed by the system.

Formula used :

$q=n\times c_v\times \Delta T$

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.773 mol

$\Delta T$ = Change in temperature = 20.1 K

$c_v$ = heat capacity at constant volume of Xe (mono-atomic molecule) = $\frac{3}{2}R$

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

$q=0.773mol\times \frac{3}{2}\times 8.314J/mol.K\times 20.1K$

$q=193.8J$

Now we have to calculate the change in internal energy of the system.

$\Delta U=q+w$

As we know that, work done is zero at constant volume. So,

$\Delta U=q=193.8J$

Therefore, the value of $q\text{ and }\Delta U$ is 193.8 J and 193.8 J respectively.