Answer:
A) 140 k 
b ) 5.22 *10^3 J
c) 2910 Pa 
Explanation:
Volume of Monatomic ideal gas = 1.20 m^3
heat added ( Q ) = 5.22*10^3 J 
number of moles  (n)  = 3 
A ) calculate the change in temp of the gas 
since the volume of gas is constant no work is said to be done 
heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT
make ΔT subject of the equation
ΔT = Q / n.(3/2).R 
     = (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )
     = 140 K 
B) Calculate the change in its internal energy 
ΔU = Q  this is because no work is done 
therefore the change in internal energy = 5.22 * 10^3 J
C ) calculate the change in pressure 
applying ideal gas equation 
P = nRT/V 
therefore ; Δ P = ( n*R*ΔT/V ) 
                         = ( 3 * 8.3144 * 140 ) / 1.20 
                         = 2910 Pa