Answer to Question #177505 in Molecular Physics | Thermodynamics for Bahaa

Question

Tw o mole s o f a monatomi c idea l ga s ar e a t a temperatur e o f 

300 K. Th e ga s expand s reversibl y an d isothermall y t o twic e it s 

original volume . Calculat e th e wor k don e b y th e gas , th e hea t 

supplied an d th e chang e in the interna l energy . 

Answer 1

Let us write the ideal gas equation for the initial and final conditions:

$p_1V_1 = \nu RT, \\ p_2V_2 = \nu RT.$

$V_2 = 2V_1 \Rightarrow p_2 = 0.5p_1.$

For the isothermal process the work done by the gas may be calculated as

$A = \nu RT\ln\dfrac{V_2}{V_1} = \nu RT\ln 2 = 2\cdot 8.31\cdot 300\cdot \ln 2 = 3456\,\mathrm{J}.$

In the isothermal process the change of temperature is 0, so the change of the internal energy is $\Delta U = \dfrac32\nu R \Delta T = 0.$

Due to the First law of thermodynamics, $\Delta U = Q - A, \; 0 = Q - 3456\,\mathrm{J},$ so the heat is $Q=3456\,\mathrm{J}.$