The percentage of heat supplied to a diatomic | vedclass

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

Question

(C) For an isobaric process, the pressure remains constant $(P = 	ext{constant})$.From the first law of thermodynamics, $\Delta Q = \Delta U + \Delta

Vedclass · Accepted Answer

$28.6$

Vedclass · Answer

(C) For an isobaric process, the pressure remains constant $(P = 	ext{constant})$.From the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$.For a diatomic gas, the molar heat capacities are $C_p = \frac{7}{2}R$ and $C_v = \frac{5}{2}R$.The heat supplied is $\Delta Q = n C_p \Delta T$.The work done is $\Delta W = n R \Delta T$.The fraction of heat converted into work is $\frac{\Delta W}{\Delta Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p} = \frac{R}{\frac{7}{2}R} = \frac{2}{7}$.Converting this to a percentage: $\frac{2}{7} 	imes 100 \approx 28.6 \%$.Therefore, the percentage of heat converted into work is $28.6 \%$.

The percentage of heat supplied to a diatomic ideal gas that is converted into work in an isobaric process is

Similar Questions

Which of the following relations is correct for an isochoric process?

When heat is given to a gas in an isobaric process,then

An ideal gas undergoes the process $1 \rightarrow 2$ as shown in the figure. The heat supplied and work done in the process are $\Delta Q$ and $\Delta W$ respectively. The ratio $\Delta Q : \Delta W$ is

In a given process for an ideal gas,$dW = 0$ and $dQ < 0$. Then for the gas:

When $1 \, mol$ of nitrogen gas is heated at constant pressure at room temperature,it is supplied with $1163.4 \, J$ of energy. The increase in temperature is ....... $K$ $(R = 8.31 \, J \, mol^{-1} \, K^{-1})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module