The amount of heat needed to raise the temperature of 4 moles of a rigid diatomic gas from $0^{\circ }\mathrm{C}$ to ${50}^{\circ }\mathrm{C}$ when no work is done is $.......\times 100\mathrm{R}.$ (R is the universal gas constant)

According to first law of thermodynamics 
$\mathrm{\Delta Q}=\mathrm{\Delta U}+\mathrm{\Delta W}$
Where, $\mathrm{\Delta Q}=$quantity of heat energy supplied to the system.
$\mathrm{\Delta U}=$change in the internal energy of a closed system.
And $\mathrm{\Delta W}=$ work done by the system on its surroundings.
As per question no work is done
$\mathrm{\Delta W}=0$

From Eqs (i) and (ii) we get 

$\begin{array}{l}\mathrm{\Delta Q}=0+\mathrm{\Delta U}\Rightarrow \mathrm{\Delta Q}=\mathrm{\Delta U}\\ \mathrm{\Delta Q}=\mathrm{\Delta U}={\mathrm{nC}}_{\mathrm{V}}\mathrm{\Delta T}\end{array}$

Or Where, C_V = specific heat capacity at constant 
volume for diatomic gas $=\frac{5\mathrm{R}}{2}\mathrm{\Delta T}=$change in temperature 
$\Rightarrow \mathrm{\Delta Q}={\mathrm{nC}}_{\mathrm{V}}\mathrm{\Delta T}$

$=(50-0)={50}^{\circ }C$

$n=\text{ number of moles }=4$

$=4\times \frac{5\mathrm{R}}{2}\times \left(50\right)=500\mathrm{R}=500\mathrm{R}$