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$N$ moles of an ideal diatomic gas are in a cylinder at temperature $T$. suppose on supplying heat to the gas, its temperature remain constant but $n$ moles get dissociated into atoms. Heat supplied to the gas is
Zero
$\frac{1}{2}nRT$
$\frac{3}{2}nRT$
$\frac{3}{2}(N - n)RT$
Solution
(b) Since the gas is enclosed in a vessel, therefore, during heating process, volume of the gas remains constant. Hence, no work is done by the gas. It means heat supplied to the gas is used to increase its internal energy only.
Initial internal energy of the gas is ${U_1} = N\,\left( {\frac{5}{2}R} \right)\,T$
Since $n$ moles get dissociated into atoms, therefore, after heating, vessel contains $(N – n)$ moles of diatomic gas and $2n$ moles of a mono-atomic gas. Hence the internal energy for the gas, after heating, will be equal to
${U_2} = (N – n)\left( {\frac{5}{2}R} \right)\,T + 2n\,\left( {\frac{3}{2}R} \right)\,T$$ = \frac{5}{2}\,NRT + \,\frac{1}{2}nRT$
Hence, the heat supplied = increase in internal energy
$ = \,({U_2} – {U_1})\, = \,\frac{1}{2}\,nRT$
Similar Questions
In Column $-I$ processes and in Column $-II$ formulas of work are given. Match them appropriately :
| Column $-I$ | Column $-II$ |
| $(a)$ Isothermal process | $(i)$ $W = \frac{{\mu R({T_1} – {T_2})}}{{\gamma – 1}}$ |
| $(b)$ Adiabatic process | $(ii)$ $W = P\Delta V$ |
| $(iii)$ $W = 2.303\,\mu RT\log \left( {\frac{{{V_2}}}{{{V_1}}}} \right)$ |
