Question

Change in enthalpy when $11.2d{{m}^{3}}$ of $He$ at NTP is heated in a cylinder to ${{100}^{0}}C$ is (assume ideal behaviour)

Answer 1

The main answer to this question lies in the calculation of enthalpy change for a monatomic gas which is given by $\Delta H=n{{C}_{p}}\Delta T$and for a monatomic gas the value of ${{C}_{p}}=\dfrac{5}{2}R$

Complete step by step solution:
In our previous classes we have come across the concept of specific heat capacity of a substance in our physical chemistry part.
- Now, let us know the main criteria about it which includes the fact that specific heat capacity of a substance is measured as the ratio of heat capacity of sample of substance to that of mass of the sample.
- Specific heat capacity is denoted by the symbol${{C}_{p}}$ and this value varies with the temperature and is different for each state of matter.
- Now, the change in the enthalpy of a system as a function of temperature and heat capacity is given by the formula,
$\Delta H=n{{C}_{p}}\Delta T$
where,$\Delta H$ is the change in enthalpy of the system
n= number of moles of substance
${{C}_{p}}$ is the specific heat capacity
and $\Delta T$is the change in the temperature of the system
Since, in the question it is given as the system behaves ideally, the specific heat capacity of a monatomic gas that is Helium is given by the formula,
${{C}_{p}}=\dfrac{5}{2}R$
where, R is the real gas constant
According to the data given we have,
At NTP , the initial temperature is 273K
and the final temperature is ${{100}^{0}}C$=373K
Thus, $\Delta T$=373-273=100K
and $R=8.314Jmo{{l}^{-1}}{{K}^{-1}}$
Number of moles of gas is the ratio of given volume to 22.4L [As per NTP conditions]
Therefore,$n=\dfrac{11.2}{22.4}=0.5moles$
Thus substituting all these values in the enthalpy equation we get,
$\Delta H=0.2\times \dfrac{5}{2}\times 8.314\times 100$
$\Rightarrow \Delta H=1039.25J$

Thus, the correct answer is $\Delta H=1039.25J$

Note: Note that specific heat of a substance, especially gas is significantly higher when it is allowed to expand that is when heated compared to that when heated in a closed vessel (here cylinder) which prevents expansion and which has a constant volume. This will lead you to the correct guesses of answer.