50 , cal of heat is required to raise the tem | vedclass

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

Question

(C) Given: Heat at constant pressure $Q_P = 50 \, cal$,number of moles $n = 1 \, mol$,change in temperature $\Delta T = 25^{\circ} C - 20^{\circ} C =

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) Given: Heat at constant pressure $Q_P = 50 \, cal$,number of moles $n = 1 \, mol$,change in temperature $\Delta T = 25^{\circ} C - 20^{\circ} C = 5 \, K$,and gas constant $R = 2 \, cal / mol \cdot K$.First,we calculate the molar heat capacity at constant pressure $(C_P)$:$Q_P = n C_P \Delta T$$50 = 1 	imes C_P 	imes 5$$C_P = 10 \, cal / mol \cdot K$.Using Mayer's relation,$C_P - C_V = R$,we find the molar heat capacity at constant volume $(C_V)$:$C_V = C_P - R$$C_V = 10 - 2 = 8 \, cal / mol \cdot K$.Now,calculate the heat required at constant volume $(Q_V)$:$Q_V = n C_V \Delta T$$Q_V = 1 	imes 8 	imes 5 = 40 \, cal$.Therefore,the correct option is $C$.

Similar Questions

The specific heat capacity of a monatomic gas at constant volume is $x \%$ of its specific heat capacity at constant pressure. Then $x=$

At constant volume,the specific heat of a gas is $\frac{3R}{2}$. Then the value of $\gamma$ will be ....

What is the relationship between the internal energy $U$ of an ideal gas and its pressure $P$ and volume $V$?

If the degree of freedom of a gas is $f,$ then the ratio of two specific heats ${C_P}/{C_V}$ is given by

The molar specific heat of a gas as given from the kinetic theory is $\frac{5}{2} R$. If it is not specified whether it is $C_P$ or $C_V$,one could conclude that the molecules of the gas

Vedclass Test Series

Exam Paper Generator

Online Exam Module