Answer:
Explanation:
Given
no of moles [tex]n=2.43[/tex]
Temperature raised [tex]\Delta T=11.9 k[/tex]
Work done by gas
[tex]W=\int_{V_1}^{V_2}PdV[/tex]
[tex]W=P\Delta V[/tex]
[tex]W=nR\Delta T[/tex]
[tex]W=2.43\times 8.314\times 11.9[/tex]
[tex]W=240.41 kJ[/tex]
(b)Energy Transferred as heat
[tex]Q=nc_p\Delta T[/tex]
[tex]c_p[/tex]=specific heat at constant Pressure
[tex]c_p[/tex] for ideal Mono atomic gas is [tex]\frac{5R}{2}[/tex]
[tex]Q=2.43\times \frac{5R}{2}\times 11.9[/tex]
[tex]Q=601.03 kJ[/tex]
(c)Change in Internal Energy
[tex]\Delta U=Q-W[/tex]
[tex]\Delta U=601.03-240.41=360.62 kJ[/tex]
(d)Change in average kinetic Energy [tex]\Delta k[/tex]
[tex]K.E._{avg}=\frac{3}{2} \times k\times T[/tex]
[tex]\Delta K.E.=\frac{3}{2} \times k\times \Delta T[/tex] ,where k=boltzmann constant
[tex]\Delta K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 11.9[/tex]
[tex]\Delta K.E.=2.46\times 10^{-22} J[/tex]
