Question 509: Thermodynamics & KTG

Question

When 300 J of heat given to an ideal gas with $C_p=\frac{7}{2} R$ its temperature raises from $20^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ keeping its volume constant. The mass of the gas is (approximately) $\_\_\_\_$ g. $(\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} . \mathrm{K})$

Accepted Answer

For a constant volume process, the heat supplied ( Q ) is given by the formula :
$$ \mathrm{Q}=\mathrm{n} \mathrm{C}_{\mathrm{v}} \Delta \mathrm{~T} $$
Where :
1. $\mathrm{n}=$ number of moles of the gas.
2. $ \mathrm{C}_{\mathrm{v}}=$ molar specific heat capacity at constant volume.
3. $\Delta \mathrm{T}=$ change in temperature $50^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}=30^{\circ} \mathrm{C}=30 \mathrm{~K}$
We are given the molar specific heat at constant pressure, $C_p=\frac{7}{2} R$. Using Mayer's Relation $\left(C_p-C_v=Right)$ :
$$ C_v=C_p-R $$
$$\Rightarrow $$ $$ C_v=\frac{7}{2} R-R=\frac{5}{2} R $$
Using the formula of heat,
$$ 300=\mathrm{n} 	imes\left(\frac{5}{2} 	imes 8.314ight) 	imes 30 $$
$$ \mathrm{n}=\frac{300 	imes 2}{5 	imes 8.314 	imes 30} \approx 0.481 \mathrm{moles} $$
So, the amount of the gas is 0.481 moles. As the molar mass of the gas is not given in the question, we can't determine the mass of the gas directly.

Thermodynamics & KTG

When 300 J of heat given to an ideal gas with C_p=\frac{7}{2} R its temperature raises from 20^{\circ} \mathrm{C} to 50^{\circ} \mathrm{C} keeping its volume constant. The mass of the gas is (approximately) \_\_\_\_ g. (\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} . \mathrm{K})

Integer / Numeric Answer Type