306 J of heat is required to raise the tempe | vedclass

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

Question

(C) Given,at constant pressure heat $Q_p = 306 \ J$.Number of moles,$n = 2$.Change in temperature,$\Delta T = 35 - 25 = 10 \ K$.We know that $Q_p = n

Vedclass · Accepted Answer

$140$

Vedclass · Answer

(C) Given,at constant pressure heat $Q_p = 306 \ J$.Number of moles,$n = 2$.Change in temperature,$\Delta T = 35 - 25 = 10 \ K$.We know that $Q_p = n C_p \Delta T$.Substituting the values: $306 = 2 	imes C_p 	imes 10$.Thus,$C_p = \frac{306}{20} = 15.3 \ J \ mol^{-1} K^{-1}$.According to Mayer's relation,$C_p - C_V = R$.Using $R \approx 8.314 \ J \ mol^{-1} K^{-1}$,we get $C_V = 15.3 - 8.314 = 6.986 \ J \ mol^{-1} K^{-1}$.The heat required at constant volume is $Q_V = n C_V \Delta T$.$Q_V = 2 	imes 6.986 	imes 10 = 139.72 \ J \approx 140 \ J$.

$306 \ J$ of heat is required to raise the temperature of $2 \ moles$ of an ideal gas at constant pressure from $25^{\circ} C$ to $35^{\circ} C$. The amount of heat required to raise the temperature of the same gas through the same range at constant volume is (in $J$)

Similar Questions

What will be the molar specific heat at constant volume of an ideal gas consisting of rigid diatomic molecules?

In the case of a diatomic gas,the fraction of heat supplied at constant pressure that is used for the expansion of the gas is:

Write the value of $\gamma$ for a polyatomic gas.

The correct relation between $\gamma = \frac{C_p}{C_v}$ and temperature $T$ is:

$A$ mixture of one mole of monoatomic gas and one mole of a diatomic gas (rigid) is kept at room temperature $\left(27^{\circ} C\right)$. The ratio of the specific heat of these gases at constant volume is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module