If 3 mol of an ideal gas at 300 K expand is | vedclass

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

Question

(D) For an isothermal process of an ideal gas,the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics,$\Delta U = Q

Vedclass · Accepted Answer

$1200$

Vedclass · Answer

(D) For an isothermal process of an ideal gas,the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics,$\Delta U = Q + W$,which implies $Q = -W$.The work done against a constant external pressure is given by $W = -P_{ext} \Delta V$.Given $P_{ext} = 80 \ kPa = 80 	imes 10^3 \ Pa$ and $\Delta V = (45 - 30) \ dm^3 = 15 \ dm^3 = 15 	imes 10^{-3} \ m^3$.$W = -80 	imes 10^3 \ Pa 	imes 15 	imes 10^{-3} \ m^3 = -1200 \ J$.Since $Q = -W$,$Q = -(-1200 \ J) = 1200 \ J$.

If $3 \ mol$ of an ideal gas at $300 \ K$ expand isothermally from $30 \ dm^3$ to $45 \ dm^3$ against a constant opposing pressure of $80 \ kPa$,then the amount of heat transferred is . . . . . . $J$.

Similar Questions

$A$ system is provided with $50 \ J$ of heat and the work done on the system is $10 \ J$. What is the change in internal energy of the system in joules?

$A$ gas absorbs $150 \ J$ of heat and expands by $300 \ cm^3$ against a constant external pressure of $2 \times 10^5 \ N \ m^{-2}$. What is the change in internal energy $(\Delta U)$ of the system (in $J$)?

When a gas undergoes adiabatic expansion,it cools down because:

Derive the relationship between $\Delta H$ and $\Delta U$ for an ideal gas. Explain each term involved in the equation.

An engine absorbs heat $Q_1$ at temperature $T_1$ and heat $Q_2$ at temperature $T_2$. The work done by the engine is given as $W = Q_1 + Q_2$. Which of the following statements is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module