(A) The heat released during combustion is absorbed by the calorimeter: $q = C_V \times \Delta T$.Given $C_V = 2.5 \ kJ \ K^{-1}$ and $\Delta T = 298.

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The heat released during combustion is absorbed by the calorimeter: $q = C_V \times \Delta T$.Given $C_V = 2.5 \ kJ \ K^{-1}$ and $\Delta T = 298.45 \ K - 298.0 \ K = 0.45 \ K$.Heat released for $3.5 \ g$ of gas $= 2.5 \ kJ \ K^{-1} \times 0.45 \ K = 1.125 \ kJ$.Moles of gas $= \frac{\text{mass}}{\text{molecular weight}} = \frac{3.5 \ g}{28 \ g \ mol^{-1}} = 0.125 \ mol$.Enthalpy of combustion per mole $= \frac{\text{Heat released}}{\text{moles}} = \frac{1.125 \ kJ}{0.125 \ mol} = 9 \ kJ \ mol^{-1}$.

Class 11 Chemistry Thermodynamics Mix Examples-Thermodynamics and Thermochemistry

In a constant volume calorimeter,$3.5 \ g$ of a gas with molecular weight $28$ was burnt in excess oxygen at $298.0 \ K$. The temperature of the calorimeter was found to increase from $298.0 \ K$ to $298.45 \ K$ due to the combustion process. Given that the heat capacity of the calorimeter is $2.5 \ kJ \ K^{-1}$,the numerical value for the enthalpy of combustion of the gas in $kJ \ mol^{-1}$ is

IIT 2009 All IIT

A
$9$
B
$4$
C
$5$
D
$6$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

Thermodynamics

Subtopic

Mix Examples-Thermodynamics and Thermochemistry

Previous Year

IIT 2009 Paper All IIT years →

$10 \ g$ each of $NH_3, N_2, Cl_2$ and $H_2S$ are expanded isothermally and reversibly at the same temperature. Identify the gas that performs maximum work.

EasyMHT CET 2024

View Solution

Two liters of $N_2$ gas at $0 \, ^\circ C$ and $5 \, atm$ pressure undergoes isothermal expansion against a constant external pressure of $1 \, atm$ until the pressure of the gas becomes $1 \, atm$. If the gas is ideal,what is the work done in the expansion in $J$?

Difficult

View Solution

Calculate the enthalpy change on freezing of $1.0 \ mol$ of water at $10.0^{\circ} C$ to ice at $-10.0^{\circ} C$. Given: $\Delta_{fus} H = 6.03 \ kJ \ mol^{-1}$ at $0^{\circ} C$,$C_p [H_2 O_{(l)}] = 75.3 \ J \ mol^{-1} \ K^{-1}$,$C_p [H_2 O_{(s)}] = 36.8 \ J \ mol^{-1} \ K^{-1}$.

Medium

View Solution

When $1.0 \ g$ of solid oxalic acid $(H_2C_2O_4)$ is burned in a bomb calorimeter whose heat capacity is $8.75 \ kJ/K$,the temperature increases by $0.312 \ K$. The enthalpy of combustion of oxalic acid at $27 \ ^oC$ is

Difficult

View Solution

Assertion : The increase in internal energy $(\Delta E)$ for the vaporization of one mole of water at $1 \ atm$ and $373 \ K$ is zero.
Reason : For all isothermal processes,$\Delta E = 0$.

EasyAIIMS 2003

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

$10 \ g$ each of $NH_3, N_2, Cl_2$ and $H_2S$ are expanded isothermally and reversibly at the same temperature. Identify the gas that performs maximum work.

Two liters of $N_2$ gas at $0 \, ^\circ C$ and $5 \, atm$ pressure undergoes isothermal expansion against a constant external pressure of $1 \, atm$ until the pressure of the gas becomes $1 \, atm$. If the gas is ideal,what is the work done in the expansion in $J$?

Calculate the enthalpy change on freezing of $1.0 \ mol$ of water at $10.0^{\circ} C$ to ice at $-10.0^{\circ} C$. Given: $\Delta_{fus} H = 6.03 \ kJ \ mol^{-1}$ at $0^{\circ} C$,$C_p [H_2 O_{(l)}] = 75.3 \ J \ mol^{-1} \ K^{-1}$,$C_p [H_2 O_{(s)}] = 36.8 \ J \ mol^{-1} \ K^{-1}$.

When $1.0 \ g$ of solid oxalic acid $(H_2C_2O_4)$ is burned in a bomb calorimeter whose heat capacity is $8.75 \ kJ/K$,the temperature increases by $0.312 \ K$. The enthalpy of combustion of oxalic acid at $27 \ ^oC$ is

Assertion : The increase in internal energy $(\Delta E)$ for the vaporization of one mole of water at $1 \ atm$ and $373 \ K$ is zero.Reason : For all isothermal processes,$\Delta E = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion : The increase in internal energy $(\Delta E)$ for the vaporization of one mole of water at $1 \ atm$ and $373 \ K$ is zero.
Reason : For all isothermal processes,$\Delta E = 0$.