Calculate work done if 1 mole of an ideal ga | vedclass

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

Question

(A) The work done in an irreversible isothermal expansion against constant external pressure is given by the formula: $W = -P_{ext} 	imes \Delta V$.

Vedclass · Accepted Answer

$-81.04$

Vedclass · Answer

(A) The work done in an irreversible isothermal expansion against constant external pressure is given by the formula: $W = -P_{ext} 	imes \Delta V$. Given: $P_{ext} = 1 \ atm$ $V_1 = 2 \ dm^3$ $V_2 = 2.8 \ dm^3$ $\Delta V = V_2 - V_1 = 2.8 \ dm^3 - 2 \ dm^3 = 0.8 \ dm^3$. Since $1 \ dm^3 \cdot atm = 101.325 \ J$,$W = -1 \ atm 	imes 0.8 \ dm^3 = -0.8 \ dm^3 \cdot atm$. $W = -0.8 	imes 101.325 \ J = -81.06 \ J$. Rounding to the nearest option,the work done is $-81.04 \ J$.

Calculate work done if $1 \ mole$ of an ideal gas expands isothermally from $2 \ dm^3$ to $2.8 \ dm^3$ against constant external pressure $1 \ atm$. (in $J$)

Similar Questions

For $1 \, \text{mole}$ of an ideal gas,isothermal expansion occurs from a volume of $2 \, L$ to $6 \, L$ at a temperature of $300 \, K$. The value of $\Delta U$ for this process is .............

Explain: Enthalpy as a state function.

For a thermodynamics system,which among the following is not a state function?

Which among the following is $NOT$ an intensive property?

$A$ system goes from $A$ to $B$ via two processes $I$ and $II$ as shown in the figure. If $\Delta U_1$ and $\Delta U_2$ are the changes in internal energies in the processes $I$ and $II$ respectively,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module