When 40 J of heat is absorbed by a monatomic | vedclass

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

Question

(C) For a monatomic gas,the degrees of freedom $f = 3$.The molar heat capacity at constant volume is $C_v = \frac{f}{2}R = \frac{3}{2}R$.The molar hea

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(C) For a monatomic gas,the degrees of freedom $f = 3$.The molar heat capacity at constant volume is $C_v = \frac{f}{2}R = \frac{3}{2}R$.The molar heat capacity at constant pressure is $C_p = C_v + R = \frac{5}{2}R$.The heat absorbed at constant pressure is $Q = n C_p \Delta T = 40 \ J$.The change in internal energy is $\Delta U = n C_v \Delta T$.Taking the ratio,$\frac{\Delta U}{Q} = \frac{n C_v \Delta T}{n C_p \Delta T} = \frac{C_v}{C_p} = \frac{3/2 R}{5/2 R} = \frac{3}{5}$.Therefore,$\Delta U = \frac{3}{5} 	imes Q = \frac{3}{5} 	imes 40 \ J = 24 \ J$.

When $40 \ J$ of heat is absorbed by a monatomic gas,the increase in the internal energy of the gas is (in $J$)

Similar Questions

Which of the following statements is correct?

If $\frac{R}{C_v} = 0.67$,identify the gas.

The relation $C_p - C_V = R$ ($C_p$ and $C_V$ are the molar specific heats at constant pressure and volume) is exactly true for

$Assertion :$ The ratio of $\frac{C_p}{C_v}$ for an ideal diatomic gas is less than that for an ideal monoatomic gas (where $C_p$ and $C_v$ have usual meaning).
$Reason :$ The atoms of a monoatomic gas have less degrees of freedom as compared to molecules of the diatomic gas.

An ideal mono-atomic gas is taken through a process such that $dQ = 3dU$. The molar heat capacity for this process is: (in $R$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module