$A$ gas expands from a volume of $4 \ L$ to $14 \ L$ against a constant external pressure of $1 \ atm$ in a cylinder with a frictionless piston. During this process,the gas absorbs $800 \ J$ of heat from the surroundings. Calculate the change in internal energy $\Delta E$ in $J$. (in $.7$)
- A$-213$
- B$-210$
- C$-217$
- D$-242$
Explore More
Similar Questions
During the melting of an ice slab at $273 \ K$ and $1 \ atm$ pressure:
Medium
View SolutionCalculate the final volume when $2$ moles of an ideal gas expand from $3 \ dm^3$ at constant external pressure $1.6 \ bar$ and the work done in the process is $800 \ J$. (in $dm^3$)
For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the relationship between $\Delta H$ and $\Delta U$ is:
Easy
View SolutionCalculate $\Delta H$ for the following reaction at $25^{\circ} C$.
$NH_2CN_{(g)} + \frac{3}{2} O_{2_{(g)}} \longrightarrow N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$
$(\Delta U = -740.5 \ kJ, R = 8.314 \ J \ K^{-1} \ mol^{-1})$
$NH_2CN_{(g)} + \frac{3}{2} O_{2_{(g)}} \longrightarrow N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$
$(\Delta U = -740.5 \ kJ, R = 8.314 \ J \ K^{-1} \ mol^{-1})$
For the reaction,$N_{2(g)} + 3 H_{2(g)} \rightarrow 2 NH_{3(g)}$,$\Delta H$ is equal to
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo