The temperature of 1 mole of an ideal monoatomic gas is increased by 50^circC$50^{\circ}C$ at constant pressure. The total heat added and change in internal energy are E_1$E_{1}$ and E_2$E_{2}$, respectively. If fracE_1E_2=fracx9$\frac{E_{1}}{E_{2}}=\frac{x}{9}$ then the value of x is
Numerical Answer Type:
Enter a numerical valueAnswer: 15 to 15+4 marks
Solution & Explanation
### Related Formula
For an ideal gas thermodynamic process:
* Total heat added at constant pressure (isobaric process) is :
E_1 = n C_P Delta T$$E_{1} = n C_{P} \Delta T$$
* Total change in internal energy is given by :
E_2 = n C_V Delta T$$E_{2} = n C_{V} \Delta T$$
The ratio of specific heat capacities is defined as:
gamma = fracC_PC_V $$\gamma = \frac{C_{P}}{C_{V}} $$
### Core Logic
Taking the ratio of the two energy expressions[cite: 179, 823]:
fracE_1E_2 = fracn C_P Delta Tn C_V Delta T = fracC_PC_V = gamma$$\frac{E_{1}}{E_{2}} = \frac{n C_{P} \Delta T}{n C_{V} \Delta T} = \frac{C_{P}}{C_{V}} = \gamma$$
### Step 1: Evaluating for a Monoatomic Gas
For an ideal monoatomic gas, the degrees of freedom are f = 3$f = 3$. This gives an adiabatic index of :
gamma = 1 + frac2f = 1 + frac23 = frac53 $$\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3} $$
Equating this value to the given ratio expression [cite: 179, 826]:
frac53 = fracx9$$\frac{5}{3} = \frac{x}{9}$$x = frac5 times 93 = 15$$x = \frac{5 \times 9}{3} = 15$$
### Pattern Recognition
The ratio of heat added to the change in internal energy during an isobaric process is always equal to the adiabatic exponent gamma$\gamma$ of the gas.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Keywords:#total heat added and change in internal energy#JEE Main 2025 Morning Q25#Thermodynamics JEE Main 2025#Specific Heat Capacities of Gases JEE Main 2025
More Thermodynamics Previous-Year Questions — Page 6
Q38jee_main_2024_31_jan_morningIsobaric Process
The given figure represents two isobaric processes for the same mass of an ideal gas, then
A Volume vs Temperature (V-T) graph showing two straight lines starting from the origin representing distinct constant pressures P1 and P2.
### Related Formula
PV = nRT$PV = nRT$
### Core Logic
From the Ideal Gas Law:
V = left(fracnRPright) T$$V = \left(\frac{nR}{P}\right) T$$
In a V-T$V-T$ graph, the equation of the line represents y = mx$y = mx$, where the slope m$m$ is:
textSlope = fracnRP$$\text{Slope} = \frac{nR}{P}$$textSlope propto frac1P$$\text{Slope} \propto \frac{1}{P}$$
Thus, a higher slope corresponds to a lower pressure.
### Step 2: Compare Slopes
From the given figure, the slope of line 2 is greater than the slope of line 1:
(textSlope)_2 > (textSlope)_1$$(\text{Slope})_2 > (\text{Slope})_1$$
Therefore, inversely:
P_2 < P_1$P_2 < P_1$
or P_1 > P_2$P_1 > P_2$.
### Pattern Recognition
In V-T graphs, steeper lines mean lower Pressure. In P-T graphs, steeper lines mean lower Volume. It's an inverse inverse slope relationship.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
More Thermodynamics Questions — jee_main_2025_24_jan_morning
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