When 1 text{ mole} of an ideal gas (gamma = 1 | vedclass

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

Question

(B) The change in internal energy $(\Delta U)$ for an ideal gas is given by the formula: $\Delta U = n C_v \Delta T$. For an ideal gas,$C_v = \frac{R}

Vedclass · Accepted Answer

$166$

Vedclass · Answer

(B) The change in internal energy $(\Delta U)$ for an ideal gas is given by the formula: $\Delta U = n C_v \Delta T$. For an ideal gas,$C_v = \frac{R}{\gamma - 1}$. Given: $n = 1 	ext{ mole}$,$\gamma = 1.4$,$R = 8.3 	ext{ J/mol K}$,$T_1 = 27^{\circ}C$,$T_2 = 35^{\circ}C$. Change in temperature $\Delta T = T_2 - T_1 = 35 - 27 = 8 	ext{ K}$. Substituting the values: $\Delta U = 1 	imes \frac{8.3}{1.4 - 1} 	imes 8$. $\Delta U = \frac{8.3}{0.4} 	imes 8 = 8.3 	imes 20 = 166 	ext{ J}$. Thus,the increase in internal energy is $166 	ext{ J}$.

When $1 \text{ mole}$ of an ideal gas $(\gamma = 1.4)$ is compressed adiabatically,its temperature increases from $27^{\circ}C$ to $35^{\circ}C$. The increase in the internal energy of the gas is ...... $J$? $(R = 8.3 \text{ J/mol K})$

Similar Questions

The relation between absolute temperature $(T)$ and pressure $(P)$ of a gas in an adiabatic process is

$A$ diatomic gas initially at $18^oC$ is compressed adiabatically to one-eighth of its original volume. The temperature after compression will be

What is the name of the ideal-gas process in which no heat is transferred?

$5.6 \text{ liter}$ of helium gas at $STP$ is adiabatically compressed to $0.7 \text{ liter}$. Taking the initial temperature to be $T_1$,the work done in the process is

What is constant in an adiabatic process?

Vedclass Test Series

Exam Paper Generator

Online Exam Module