An ideal gas is expanded from 1 times 10^{-3} | vedclass

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

Question

(D) The work done during expansion against constant external pressure is given by the formula: $W = -P_{ex} \Delta V$. Given: $P_{ex} = 1 	imes 10^5

Vedclass · Accepted Answer

$-900 \,J$

Vedclass · Answer

(D) The work done during expansion against constant external pressure is given by the formula: $W = -P_{ex} \Delta V$. Given: $P_{ex} = 1 	imes 10^5 \,N/m^2$ $V_1 = 1 	imes 10^{-3} \,m^3$ $V_2 = 1 	imes 10^{-2} \,m^3 = 10 	imes 10^{-3} \,m^3$ Change in volume: $\Delta V = V_2 - V_1 = (10 	imes 10^{-3} - 1 	imes 10^{-3}) \,m^3 = 9 	imes 10^{-3} \,m^3$ Calculating work: $W = -(1 	imes 10^5 \,N/m^2) 	imes (9 	imes 10^{-3} \,m^3)$ $W = -9 	imes 10^2 \,J = -900 \,J$

An ideal gas is expanded from $1 \times 10^{-3} \,m^3$ to $1 \times 10^{-2} \,m^3$ against a constant external pressure of $1 \times 10^5 \,N/m^2$ at $300 \,K$. The work done is

Similar Questions

Identify the process having $\Delta V = 0$.

Assertion : Absolute values of internal energy of substances cannot be determined.
Reason : It is impossible to determine exact values of constituent energies of the substances.

In thermodynamics,the $P-V$ work done is given by $w = -\int P_{ext} dV$. For a system undergoing a particular process,the work done is $w = -\int \left(\frac{RT}{V - b} - \frac{a}{V^2}\right) dV$. This equation is applicable to a:

When $2 \ L$ of an ideal gas expands isothermally into a vacuum to a total volume of $6 \ L$,what is the change in internal energy in $J$? (Nearest integer)

Internal energy is

Vedclass Test Series

Exam Paper Generator

Online Exam Module