0.08text kg$0.08\text{ kg}$ of air is heated at constant volume through 5^circtextC$5^{\circ}\text{C}$. The specific heat of air at constant volume is 0.17text kcal/kg^circtextC$0.17\text{ kcal/kg}^{\circ}\text{C}$ and J = 4.18text joule/cal$J = 4.18\text{ joule/cal}$. The change in its internal energy is approximately:
A.318text J$318\text{ J}$
B.298text J$298\text{ J}$
C.284text J$284\text{ J}$
D.142text J$142\text{ J}$
Solution & Explanation
### Related Formula
Q = Delta U + W$$Q = \Delta U + W$$
Since the volume is constant, work done W = 0$W = 0$, leading to:
Delta U = Q = m c_v Delta T$$\Delta U = Q = m c_v \Delta T$$
### Core Logic
Given values:
m = 0.08text kg$m = 0.08\text{ kg}$c_v = 0.17text kcal/kg^circtextC = 0.17 times 10^3text cal/kg^circtextC$c_v = 0.17\text{ kcal/kg}^{\circ}\text{C} = 0.17 \times 10^3\text{ cal/kg}^{\circ}\text{C}$Delta T = 5^circtextC$\Delta T = 5^{\circ}\text{C}$J = 4.18text J/cal$J = 4.18\text{ J/cal}$
### Step 1: Metric conversion and computing value
Delta U = 0.08 times (0.17 times 10^3) times 5 times 4.18$$\Delta U = 0.08 \times (0.17 \times 10^3) \times 5 \times 4.18$$Delta U = 0.08 times 170 times 5 times 4.18$$\Delta U = 0.08 \times 170 \times 5 \times 4.18$$Delta U = 68 times 4.18 simeq 284.24text J approx 284text J$$\Delta U = 68 \times 4.18 \simeq 284.24\text{ J} \approx 284\text{ J}$$
### Pattern Recognition
Isochoric processes channel total thermal input strictly into core state configurations, converting raw metric calories directly to standard mechanical Joules via mechanical equivalent factors (J$J$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Keywords:#air is heated at constant volume#JEE Main 2024 Morning Q40#Thermodynamics JEE Main 2024#First Law of Thermodynamics JEE Main 2024
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_2024_27_jan_morning
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