When the temperature of 2 moles of an ideal g | vedclass

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

Question

(B) For an ideal gas,the equation of state is $PV = nRT$. Since the pressure $P$ is constant,the change in volume $\Delta V$ due to a change in temper

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) For an ideal gas,the equation of state is $PV = nRT$. Since the pressure $P$ is constant,the change in volume $\Delta V$ due to a change in temperature $\Delta T$ is given by $P \Delta V = nR \Delta T$. The work done $W$ in a process at constant pressure is defined as $W = P \Delta V$. Substituting the expression for $P \Delta V$,we get $W = nR \Delta T$. Given $n = 2$ moles and $\Delta T = 20 \ K$ (or $20^{\circ} C$),the work done is $W = 2 	imes R 	imes 20 = 40R$.

When the temperature of $2$ moles of an ideal gas is increased by $20^{\circ} C$ at constant pressure,find the work involved in the process. (in $R$)

Similar Questions

Which among the following pairs represents extensive and intensive properties respectively?

$6$ moles of an ideal gas expand isothermally and reversibly from a volume of $1 \ L$ to a volume of $10 \ L$ at $27 \ ^oC$. What is the maximum work done? (in $kJ$)

It is a general principle that the less energy a system contains,it is:

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

$6$ moles of an ideal gas expand isothermally and reversibly from a volume of $1$ litre to a volume of $10$ litres at $27\,^oC$. What is the maximum work done (in $, kJ$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module