If the temperature of a gas is increased from | vedclass

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

Question

(D) The average kinetic energy $(E)$ of an ideal gas is directly proportional to its absolute temperature $(T)$,given by the relation $E = \frac{3}{2}

Vedclass · Accepted Answer

Four times

Vedclass · Answer

(D) The average kinetic energy $(E)$ of an ideal gas is directly proportional to its absolute temperature $(T)$,given by the relation $E = \frac{3}{2} k_B T$.Therefore,the ratio of kinetic energies is $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.Given temperatures in Celsius: $T_1 = 27^{\circ}C$ and $T_2 = 927^{\circ}C$.Converting to Kelvin: $T_1 = 27 + 273 = 300 \ K$ and $T_2 = 927 + 273 = 1200 \ K$.Substituting the values: $\frac{E_1}{E_2} = \frac{300}{1200} = \frac{1}{4}$.Thus,$E_2 = 4 E_1$,which means the kinetic energy becomes four times the initial value.

If the temperature of a gas is increased from $27^{\circ}C$ to $927^{\circ}C$,what will be the change in its kinetic energy?

Similar Questions

The total momentum of the molecules of $1 \,gm$ $mol$ of a gas in a container at rest at $300 \,K$ is

The temperature at which the average translational kinetic energy of a molecule is equal to the energy gained by an electron in accelerating from rest through a potential difference of $1\, V$ is:

The temperature of a gas is $-73^{\circ}C$. To what temperature in $^{\circ}C$ should the gas be heated so that the average kinetic energy of its molecules becomes doubled?

The pressure exerted by an ideal gas at a particular temperature is directly proportional to

The kinetic energy of one gm-mole of a gas at normal temperature and pressure is $(R = 8.31 \text{ J/mol-K})$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module