Answer
a) $62.3\;\rm J$
b) $104\;\rm J$
c) $145\;\rm J$
Work Step by Step
Since the container is rigid, then the gas is heated with constant volume. This means that the work done on the gas is zero and hence the heat needed is given by
$$Q=\Delta E_{th}=nC_{\rm v}\Delta T$$
$$Q= nC_{\rm v}\Delta T\tag 1$$
Now we need to find the number of moles of the gas, which is given by
$$n=\dfrac{m_{H_2}}{M_{H_2}}=\dfrac{m_{H_2}}{2M_{H}}$$
Plugging the known;
$$n =\dfrac{ 0.20}{2(1) }=\bf 0.10\;\rm mol$$
Plugging into (1);
$$Q=0.10C_{\rm v}\Delta T\tag 2$$
Noting that:
A- From 0 K to 100 K, $C_{\rm v}=\frac{3}{2}R$,
B- From 200 K to 1000 K $C_{\rm v}=\frac{5}{2}R$,
C- From 2000 K to 10000 K $C_{\rm v}=\frac{7}{2}R$,
See Figure 18.13 of Hydrogen molar specific heat at constant volume as a function of temperature.
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a) Using the formula (2) from above, and case A for $C_{\rm v}$:
$$Q=0.10\left( \frac{3}{2}R\right) (100-50)=0.10\left( \frac{3}{2}\times 8.31\right) (100-50)$$
$$Q=\color{red}{\bf 62.3}\;\rm J$$
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b) Using the formula (2) from above, and case B for $C_{\rm v}$:
$$Q=0.10\left( \frac{5}{2}R\right) (300-250)=0.10 \left( \frac{5}{2}\times 8.31\right) (300-250)$$
$$Q=\color{red}{\bf 104}\;\rm J$$
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c) Using the formula (2) from above, and case C for $C_{\rm v}$:
$$Q=0.10\left( \frac{7}{2}R\right) (2300-2250)=0.10 \left( \frac{7}{2}\times 8.31\right) (2300-2250)$$
$$Q=\color{red}{\bf 145}\;\rm J$$
