10 mole of oxygen is heated at constant volume from 30^circtextC$30^{\circ}\text{C}$ to 40^circtextC$40^{\circ}\text{C}$. The change in the internal energy of the gas is ____ cal. (The molecular specific heat of oxygen at constant pressure, C_p = 7text cal./mol ^circtextC$C_{p} = 7\text{ cal./mol }^{\circ}\text{C}$ and R = 2text cal./mol ^circtextC$R = 2\text{ cal./mol }^{\circ}\text{C}$.)
Numerical Answer Type:
Enter a numerical valueAnswer: 500 to 500+4 marks
Solution & Explanation
### Related Formula
Delta U = n C_v Delta T$$\Delta U = n C_v \Delta T$$C_v = C_p - R$C_v = C_p - R$
### Core Logic
Given values:
n = 10text moles$n = 10\text{ moles}$Delta T = 40^circtextC - 30^circtextC = 10^circtextC$\Delta T = 40^{\circ}\text{C} - 30^{\circ}\text{C} = 10^{\circ}\text{C}$C_p = 7text cal/molcdot^circtextC$C_p = 7\text{ cal/mol}\cdot^{\circ}\text{C}$R = 2text cal/molcdot^circtextC$R = 2\text{ cal/mol}\cdot^{\circ}\text{C}$
First, find C_v$C_v$ using Mayer's relation:
C_v = C_p - R = 7 - 2 = 5text cal/molcdot^circtextC$$C_v = C_p - R = 7 - 2 = 5\text{ cal/mol}\cdot^{\circ}\text{C}$$
### Step 1: Calculate Internal Energy Change
Delta U = n C_v Delta T$$\Delta U = n C_v \Delta T$$Delta U = 10 times (7 - 2) times (40 - 30)$$\Delta U = 10 \times (7 - 2) \times (40 - 30)$$Delta U = 10 times 5 times 10 = 500text cal$$\Delta U = 10 \times 5 \times 10 = 500\text{ cal}$$
### Pattern Recognition
Change in internal energy of an ideal gas is ALWAYS Delta U = n C_v Delta T$\Delta U = n C_v \Delta T$, regardless of the process (constant volume or not). Use C_v = C_p - R$C_v = C_p - R$ when C_p$C_p$ is given.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Class 11 Physics: Kinetic Theory of Gases
Keywords:#10 mole of oxygen is heated at constant volume#JEE Main 2026 Morning Q49#Thermodynamics JEE Main 2026#Internal Energy JEE Main 2026
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_2026_21_jan_morning
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