A gas is allowed to expand in an insulated co | vedclass

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

Question

(D) For an insulated container,the process is adiabatic,so $q = 0$. According to the first law of thermodynamics,$\Delta U = q + w$. The work done dur

Vedclass · Accepted Answer

$-506.5$

Vedclass · Answer

(D) For an insulated container,the process is adiabatic,so $q = 0$. According to the first law of thermodynamics,$\Delta U = q + w$. The work done during expansion against a constant external pressure is given by $w = -P_{ext} 	imes \Delta V$. Here,$P_{ext} = 2.5 \ atm$,$V_1 = 2.5 \ L$,and $V_2 = 4.5 \ L$. $\Delta V = V_2 - V_1 = 4.5 \ L - 2.5 \ L = 2.0 \ L$. $w = -2.5 \ atm 	imes 2.0 \ L = -5.0 \ atm \cdot L$. Since $1 \ atm \cdot L = 101.3 \ J$,we have $w = -5.0 	imes 101.3 \ J = -506.5 \ J$. Therefore,$\Delta U = 0 + (-506.5 \ J) = -506.5 \ J$.

$A$ gas is allowed to expand in an insulated container against a constant external pressure of $2.5 \ atm$ from $2.5 \ L$ to $4.5 \ L$. The change in internal energy of the gas in joules is: (in $J$)

Similar Questions

For the reaction $3O_{2(g)} \rightarrow 2O_{3(g)}$,the value of $\Delta U - \Delta H$ is equal to .....

For the reaction $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,the relationship between $\Delta H$ and $\Delta U$ is:

Under isothermal condition,a gas expands from $0.2 \ dm^3$ to $0.8 \ dm^3$ against a constant external pressure of $2 \ bar$ at $300 \ K$. Find the work done by the gas. (in $J$)

If $1.5 \ L$ of an ideal gas at a pressure of $20 \ atm$ expands isothermally and reversibly to a final volume of $15 \ L$,the work done by the gas in $L \ atm$ is

Calculate $\Delta U$ if $2 \ kJ$ heat is released and $10 \ kJ$ of work is done on the system. (in $kJ$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module