The kinetic energy of 0.5 mole of N2 gas is x J at 1.5 atm pressure then the kinetic energy of the gas at 4.5 atm pressure and at same temperature will be
3x J
x J
Answers
Example 1.1 The atmosphere consists of 78.08% by volume of N2, and 20.95% of O2. Calculate the
partial pressures due to the two gases.
The specification "% by volume" may be interpreted as follows. If the components of the
atmosphere were to be separated, at the pressure of 1 atm, the volume occupied by each component is
specified by the volume %. Thus, if we isolate the N2 in 1.000 L of dry air, at a pressure of 1 atm its
volume will be 0.781 L. According to the ideal gas law, at a fixed pressure and temperature, the amount
of gas in moles N = V(p/RT), i.e. the amount in moles is proportional to the volume. Hence percentage
by volume is the same as percentage in N, i.e. 1.000 moles of air consists of 0.781 moles of N2.
According to the Dalton's law (see (1.3.5)) the partial pressure is proportional to the N, the partial
pressure of N2 is 0.781 atm and that of O2 is 0.209 atm.
Example 1.2 Using the ideal gas approximation, estimate the change in the total internal energy of 1.00 L
of N2 at p=2.00 atm and T = 298.15 K, if its temperature is increased by 10.0 K. What is the energy
required to heat 1.00 mole of N2 from 0.0 K to 298 K ?
The energy of an ideal gas depends only on the amount of gas N and the temperature. For a
diatomic gas such as N2 the energy per mole equals (5/2)RT+U0. Hence, for N moles of N2 the change
in energy ΔU for a change in temperature from T1 to T2 is:
ΔU = N(5/2)R(T2 – T1)
In the above case N=pV/RT =
2.00atm x 1.00L
0.0821 L.atm.mol !1
K!1
(298.15)
= 8.17x10!2mol .
Hence:
!U = 8.17x10"2mol 5
2
8.314J.mol"1
.K"1 ( )(10.0K)
= 17.0J
(Note the different units of R used in this calculation.)
The energy required to heat 1.00 mol of N2 from 0 K to 298 K is
=(5/2)RT=(5/2)(8.314JK-1
mol-1
)298K=6.10kJ mol-1
Answer:
x J
Explanation:
KE1 = 3/2 nRT
Xj = 3/2 nRT
KE2 = 3/2 nRT
KE1= KE2
Xj = KE2
KE2 = Xj