Answer to Question #122497 in Molecular Physics | Thermodynamics for Thandeka

Question

Three moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2438 J of
heat is added to the gas, and 962 J of work is done on it. Calculate the final
temperature of the gas.

Answer 1

As per the given question,

Number of moles of the mono atomic gas=3

Initial temperature $(T_1)=345K$

Added heat =2438J

Work done on the system=-962J

Final temperature of the gas =?

We know that change in the internal energy

$\Delta U=\frac{3nR(T_2-T_1)}{2}$

$\Delta U=\frac{3\times 3\times 8.314\times (T_2-345)}{2}$

Now, we know that

$Q=\Delta U +W$

Now substituting the values,

$\Rightarrow 2438 =\frac{3\times 3\times 8.314\times (T_2- 345)}{2}-962$

$\Rightarrow (T_2-345)=\frac{2(2438+962)}{9\times 8.314}=90.88K$

$T_2= 345+90.87= 435.88K$