Question 728: Thermodynamics & KTG

Question

If heat energy $\Delta \mathrm{Q}$ is supplied to an ideal diatomic gas, the increase in internal energy is $\Delta U$ and the amount of work done by the gas is $\Delta \mathrm{W}$. The ratio $\Delta \mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q}$ is

Accepted Answer

When heat energy $\Delta Q$ is supplied to an ideal diatomic gas, the increase in internal energy is $\Delta U$, and the work done by the gas is $\Delta W$. We need to find the ratio $\Delta W : \Delta U : \Delta Q$.
The fraction of heat energy used to perform external work is determined by the formula:
$ \left(\frac{\Delta W}{\Delta Q}ight) = \left(1 - \frac{1}{\gamma}ight) $
Here, $\gamma$ (gamma) for a diatomic gas is $\frac{7}{5}$. Thus, substituting into the formula:
$ \frac{\Delta W}{\Delta Q} = 1 - \frac{1}{\left(\frac{7}{5}ight)} = 1 - \frac{5}{7} = \frac{2}{7} \quad 	ext{.....(i)} $
Next, the fraction of heat energy used to increase the internal energy is:
$ \frac{\Delta U}{\Delta Q} = \frac{1}{\gamma} = \frac{5}{7} \quad 	ext{.....(ii)} $
Therefore, combining equations (i) and (ii), the ratios are:
$ \Delta W : \Delta U : \Delta Q = 2 : 5 : 7 $

Answer

$2: 3: 5$

Answer

$2: 5: 7$

Answer

$7: 5: 9$

Answer

$1: 2: 5$

Thermodynamics & KTG

If heat energy \Delta \mathrm{Q} is supplied to an ideal diatomic gas, the increase in internal energy is \Delta U and the amount of work done by the gas is \Delta \mathrm{W}. The ratio \Delta \mathrm{W}: \Delta \mathrm{U}: \Delta \mathrm{Q} is

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