For a thermodynamic process delta Q = -50 calorie and W = -20 calorie . If initial internal energy i

English

Gujarati

Hindi

11.Thermodynamics

medium

For a thermodynamic process $\delta Q = -50$ $calorie$ and $W = -20$ $calorie$ . If the initial internal energy is $-30$ $calorie$ then final internal energy will be ....... $calorie$

A

$191.20$

B

$-60$

C

$100$

D

$-100$

Solution

$\delta \mathrm{Q}=\mathrm{W}+\delta \mathrm{U}$

$-50=-20+\left(U_{t}-U_{1}\right)$

$\mathrm{U}_{\mathrm{f}}-(-30)=-30$

$\mathrm{U}_{\mathrm{f}}=-60 \mathrm{Cal}$

Standard 11

Physics

Share

Similar Questions

A fixed amount of a gas undergoes a thermodynamic process as shown such that heat interaction along path $B \to C \to A$ is equal to the work done by the gas along path $A \to B \to C$. Then process $A \to B$ is :-

medium

A shown in the figure, the amount of heat absorbed along the path $ABC$ is $90\,J$ and the amount of work done by the system is $30\,J$ . If the amount of work done along the path $ADC$ is $20\,J$ , the amount of heat absorbed will be …. $J$

medium

The $P-V$ diagram of a diatomic ideal gas system going under cyclic process as shown in figure. The work done during an adiabatic process $CD$ is (use $\gamma=1.4$) (in $J$)

medium

(JEE MAIN-2021)

The $P-V$ graph of an ideal gas cycle is shown here as below. The adiabatic process is described by

easy

Volume versus temperature graph of two moles of helium gas is as shown in figure. The ratio of heat absorbed and the work done by the gas in process $1-2$ is

medium