Question

The kinetic energy of I kg of oxygen at 300 K
is 1 356X106 J. Find the kinetic energy of 4 kg
of oxygen at 400 K. (Ans : 7.232 x 106)​

Answer 1

$\Large\underline{\underline{\bf \red{Question}:}}$

The kinetic energy of 1 kg of oxygen at 300 K is $\sf{1.356×10^6}$ J. Find the kinetic energy of 4 kg

of oxygen at 400 K

$\Large\underline{\underline{\sf Given:}}$

• Mass of oxygen $\sf{m_1}$=1kg

• Temperature $\sf{T_1}$=300K

• Kinetic Energy $\sf{KE_1}$=$\sf{1.356×10^{6}}$

Another oxygen of

• Mass $\sf{m_2}$=4kg

• Temperature $\sf{T_2}$=400K

$\Large\underline{\underline{\sf To\:Find:}}$

• Kinetic Energy of oxygen $\sf{KE_2}$=?

$\Large\underline{\underline{\sf Formula\:Used:}}$

$\large{\boxed{\boxed{\sf \pink{KE=\dfrac{3}{2}\dfrac{mRT}{M}}}}}$

Here ,

KE = Kinetic Energy

m = mass

M = Molecular Mass

R = Gas Constant

T = Temperature

$\Large\underline{\underline{\sf Solution:}}$

$\implies{\sf \dfrac{KE_1}{KE_2}=\dfrac{\dfrac{3}{2}\dfrac{m_1RT_1}{M}}{\dfrac{3}{2}\dfrac{m_2RT_2}{M}}}$

$\implies{\sf \dfrac{KE_2}{KE_2}=\dfrac{m_1T_1}{m_2T_2} }$

$\implies{\sf \dfrac{1.356×10^6}{KE_2}=\dfrac{1×300}{4×400} }$

$\implies{\sf \dfrac{1.356×10^6×4×400}{300}=KE_2 }$

$\implies{\sf KE_2=\dfrac{21.696×10^6}{3} }$

$\implies{\sf \red{KE_2=7.232×10^6\:J}}$

$\Large\underline{\underline{\sf Answer:}}$

⛬ Kinetic Energy $\sf{(KE_2)}$ is $\sf{7.232×10^6\:J}$