An ideal gas expands isothermally from 10^{-3 | vedclass

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

Question

(B) The work done on the gas during an expansion against a constant external pressure is given by the formula: $W = -P_{ext} 	imes \Delta V$Given:$P_

Vedclass · Accepted Answer

$-900 \; J$

Vedclass · Answer

(B) The work done on the gas during an expansion against a constant external pressure is given by the formula: $W = -P_{ext} 	imes \Delta V$Given:$P_{ext} = 10^{5} \; Nm^{-2}$$V_i = 10^{-3} \; m^{3}$$V_f = 10^{-2} \; m^{3}$Change in volume $\Delta V = V_f - V_i = 10^{-2} - 10^{-3} = 10^{-2} - 0.1 	imes 10^{-2} = 0.9 	imes 10^{-2} \; m^{3}$Substituting the values:$W = -10^{5} \; Nm^{-2} 	imes (0.9 	imes 10^{-2} \; m^{3})$$W = -0.9 	imes 10^{3} \; J$$W = -900 \; J$Since the work done is negative,it indicates that work is done by the gas on the surroundings.

An ideal gas expands isothermally from $10^{-3} \; m^{3}$ to $10^{-2} \; m^{3}$ at $300 \; K$ against a constant external pressure of $10^{5} \; Nm^{-2}$. The work done on the gas is:

Similar Questions

Which of the following is a state function: $(i)$ height of a hill,$(ii)$ distance covered in climbing the hill,$(iii)$ change in energy while climbing the hill?

Explain extensive property and intensive property.

Which expression is correct for the work done in adiabatic reversible expansion of an ideal gas?

Work done when $2 \ mol$ of an ideal gas is compressed from a volume of $5 \ m^{3}$ to $2.5 \ m^{3}$ at $300 \ K$, under a constant external pressure of $100 \ kPa$ is: (in $kJ$)

During the reversible isothermal expansion of an ideal gas,the change in internal energy is......

Vedclass Test Series

Exam Paper Generator

Online Exam Module