Question

The temperature of 2.00 mol of an ideal monatomic gas is raised 15.0 K at constant volume. What are (a) the work W done by the gas, (b) the energy transferred as heat Q , (c) the change ?E_int in the internal energy of the gas, and (d) the change ?K in the average kinetic energy per atom

Answer 1

(a)
Generally work done by the gas is given by the integral
W = ? p dV from initial to final volume
For a constant pressure process it simplifies to:
W = P ? ? dV = P??V

Change in volume can be found from ideal gas law:
P?V = n?R?T
<=>
V = (n?R/P)?T
Because pressure P and number of moles n are constant for this process, change in volume and change in temperature are related as:
?V = (n?R/P)??T

Hence,
W = P??V = n?R??T
= 2mol ? 8.3145J/molK ? 15K
= 249.4 J


(b)
Heat transferred in a constant pressure process equals the change in in enthalpy. The change in enthalpy for an ideal gas is given by:
?H = n?C_p??T
Molar heat capacity at constant pressure for an monatomic ideal gas is:
C_p = (5/2)?R

Thus,
Q = ?H = (5/2)?n?R??T
= (5/2) ? 2mol ? 8.3145J/molK ? 15K
= 623.6 J


(c)
The change in internal energy equals the heat transferred to minus the work done by the gas:
?E = Q - W = (5/2)?n?R??T - n?R??T
= (3/2)?n?R??T
= (3/2) ? 2mol ? 8.3145J/molK ? 15K
= 374.2 J


(d)
In an ideal gas of gas particles of mass m at absolute temperature T the particles move at a root mean square velocity of:
v_rms = ?( 3?k_b?T/m )
(k_b is he Boltzmann constant)
So the kinetic energy of a single gas particle is given by:
K = (1/2)?m?(v_rms)