Question

Under constant pressure, the temperature of 2.00 mol of an ideal monotomic gas is raised 15.0 K. What are a) the work done by the gas b) energy transferred as heat Q c) the change Delta-Eint in the internal energy of the gas, and d) the change Delta-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)