Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main 202229 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
29 June
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2022 · 29 June · Evening Shift · Chemistry · Thermodynamics (C)

2 . 2g of nitrous oxide ( N _ 2 O) gas is cooled at a constant...

2.2g2 . 2g of nitrous oxide (N2O)\left({N}_{2}O\right) gas is cooled at a constant pressure of 11 atm from 310Kto270K310K to 270K causing the compression of the gas from 217.1mLto167.75mL217 . 1mL to 167 . 75mL. The change in internal energy of the process, △U△ U is ′−x′J' - x ' J. The value of ′x′' x ' is _____.

[nearest integer]

(Given: atomic mass of N=14gmol−1N = 14g{mol}^{- 1} and of O=16gmol−1O = 16g{mol}^{- 1}. Molar heat capacity of N2O{N}_{2}O is 100JK−1mol−1100{JK}^{- 1}{mol}^{- 1})

Correct answer

0

Step-by-step solution

Molecular weight of nitrous oxide ((N2O)=\left(\mathrm{N}_{2} \mathrm{O}\right)=44 \mathrm{ g}/ mol)Number\mathrm{mol}) Number of molesof(N2O=2.244moles of (\mathrm{N}_{2} \mathrm{O}=\frac{2.2}{44} \quad0.05 mol)Specificheatcapacity\mathrm{ mol}) Specific heat capacity of (N2O=880J/g.K)(∴)Heatreleaseoncoolingfrom(310K)to(270K)(q=−2.2×880×40K)(q=−77.44KJ(\mathrm{N}_{2} \mathrm{O}=880 \mathrm{ J}/ \mathrm{g}. \mathrm{K}) (\therefore) Heat release on cooling from (310 \mathrm{ K}) to (270 \mathrm{ K}) (q=-2.2 \times880 \times40 \mathrm{ K}) (q=-77.44 \mathrm{KJ} Negative sign indicates heat released. (\therefore) Work done by surrounding on system to compares the gas from (217.1 ml)\mathrm{ml}) to (167.75ml(167.75 \mathrm{ml} at constant pressure. It means work done by surrounding is irreversible. ∴W=P×ΔV=1.01×105×(167.75217.1mll)=1.01×105×0.049JW=4.95k.J\begin{aligned} & \therefore W= P \times \Delta V \\ & =1.01 \times10^5 \times(167.75 \quad 217.1 \mathrm{ml}l) \\ & =1.01 \times10^5 \times0.049 J \\ & W=4.95 k. J \end{aligned}) Using 1st Lawofthermodynamicsof thermodynamicsΔU=q+WΔU=−77.44KJ+4.95KJΔU=−72.49KJ\begin{aligned} & \Delta U=q+W \\ & \Delta U=-77.44 K J + 4.95 K J \\ & \Delta U=-72.49 K J \end{aligned}).<br>).<br>\therefore x = 72.$
Practice Thermodynamics (C) on Quantrex Academy →

More from Thermodynamics (C)

An ideal gas ( 0.5 mol), initially at 2 bar pressure, is compressed at a constant… JEE Advanced 2026List-I contains various physical/chemical processes, and List-II contains combinations of… JEE Advanced 2026Consider the reaction 2H_2S(g) + 3O_2(g) arrow 2H_2O(l) + 2SO_2(g) The magnitude of… JEE Main 2026Match List - I with List - II. Given V_1 and V_2 are initial and final volumes… JEE Main 2026Arrange the following isothermal processes in order of the magnitude of the work ( p - V… JEE Main 2026Consider the reaction X leftharpoons Y at 300 K. If Δ H^ θ and K are 28.40 kJ mol ^ -1… JEE Main 2026Consider the following data for the reaction X_2(g) + Y_2(g) leftharpoons 2XY(g) at 600 K… JEE Main 2026The correct order of molar heat capacities measured at 298 K and 1 bar is : JEE Main 2026 Full Thermodynamics (C) list All JEE Main PYQs