At 27° C temperature, the kinetic energy of an ideal gas is E1 . If the temperature is increased to 327° C, then kinetic energy would be

Question

At 27° C temperature, the kinetic energy of an ideal gas is E1 . If the temperature is increased to 327° C, then kinetic energy would be (A) (E1/2) (B) (E1/√2) (C) √2 E1 (D) 2 E1

Tardigrade Admin · Accepted Answer

Kinetic energy of an ideal gas, Ek = (3/2) RT or Ek propto T ∴ E327 = E27 ((327 + 273)/(27 + 273)) = 2E27 = 2E1

Q. At $2 7^{\circ} C$ temperature, the kinetic energy of an ideal gas is $E_{1} .$ If the temperature is increased to $32 7^{\circ} C$ , then kinetic energy would be

$\frac{E _{1}}{2}$

$\frac{E _{1}}{2}$

$2 E_{1}$

$2 E_{1}$

Kinetic energy of an ideal gas,
$E_{k} = \frac{3}{2} RT$ or $E_{k} \propto T$
$∴ E_{327} = E_{27} \frac{( 327 + 273 )}{( 27 + 273 )} = 2 E_{27} = 2 E_{1}$

Q. At 27∘C temperature, the kinetic energy of an ideal gas is E1​. If the temperature is increased to 327∘C, then kinetic energy would be

2E1​​

2​E1​​

2​E1​

2E1​

Kinetic energy of an ideal gas, Ek​=23​RT or Ek​∝T ∴E327​=E27​(27+273)(327+273)​=2E27​=2E1​

Q. At $2 7^{\circ} C$ temperature, the kinetic energy of an ideal gas is $E_{1} .$ If the temperature is increased to $32 7^{\circ} C$ , then kinetic energy would be

$\frac{E _{1}}{2}$

$\frac{E _{1}}{2}$

$2 E_{1}$

$2 E_{1}$

Kinetic energy of an ideal gas,
$E_{k} = \frac{3}{2} RT$ or $E_{k} \propto T$
$∴ E_{327} = E_{27} \frac{( 327 + 273 )}{( 27 + 273 )} = 2 E_{27} = 2 E_{1}$