Question

Calculate the heat required to raise the temperature of 8 grams of helium at constant volume by $20{}^\circ C$(in Cal).

Answer 1

We have given the change in temperature or rise in temperature and the mass of the gas helium. By using the first law of thermodynamics and the formula for internal energy at constant volume we can find an equation for the heat added to the system. Therefore, here we consider helium as an ideal gas.

Formula used:

& dU=Q-W \\\ & dU=n{{C}_{V}}dT \\\ & \text{number of moles, }n=\dfrac{\text{mass of the substance}}{\text{molar mass of the substance}} \\\ & {{C}_{V}}=\dfrac{fR}{2} \\\ \end{aligned}$$ **Complete answer:** We know that according to the first law of thermodynamics, change in internal heat is given as the difference between heats added to the system and the work done by the system. Mathematical representation of the first law is $$dU=Q-W$$ Now at constant volume, as there is no change in volume of the system, hence no work will be done. Therefore, at constant volume $$dU=Q\text{ }...................\text{(i)}$$ Now for constant volume internal energy is given as $$dU=n{{C}_{V}}dT\text{ }....................\text{(ii)}$$ where n is the number of moles in gas, $${{C}_{V}}$$ is the specific heat at constant volume and dT is the change in temperature. Hence from equations (i) and (ii), we can write $$Q=n{{C}_{V}}dT\text{ }..................\text{(iii)}$$ Now we got an equation to calculate heat required to raise temperature, it is given $$dT=20{}^\circ C=293.15K$$whereas formula for specific heat at constant volume and number of moles are given below $$\begin{aligned} & \text{number of moles, }n=\dfrac{\text{mass of the substance}}{\text{molar mass of the substance}} \\\ & {{C}_{V}}=\dfrac{fR}{2} \\\ \end{aligned}$$ Mass of the substance is given 8 grams and molar mass for helium gas is 4 gram/mole. In specific heat at constant volume formula f is degrees of freedom and R is universal gas constant $$R=1.986cal/moleK$$ and as helium is monoatomic gas therefore $$f=3$$. By substituting the values in the above formula for n, we get $$n=\dfrac{8}{4}=2mole$$ And substituting value of R and f for helium gas, specific heat becomes $$\begin{aligned} & {{C}_{V}}=\dfrac{3\times 1.986}{2} \\\ & \Rightarrow {{C}_{V}}=2.979cal/moleK \\\ \end{aligned}$$ Now substituting the values of n, $${{C}_{V}}$$and dT in equation (iii) we get $$\begin{aligned} & Q=2\times 2.979\times 293.15 \\\ & Q=1746.5877cal \\\ \end{aligned}$$ Hence 1746.5877calories of heat is required to raise temperature by $$20{}^\circ C$$. **Note:** The SI unit of heat is Joules, as it was given in the question itself that we have to calculate the heat in calorie therefore the final answer is in calorie. Although we changed the unit of temperature in kelvin because the specific heat’s unit has kelvin and if we didn’t change the unit of temperature then the answer would be wrong as well as the unit.