Question

. На сколько увеличится внутренняя энергия идеального идеального одноатомного газа, если 2/3 моля этого газа нагреть на 1ОО К ?

Answer 1

ΔU =3mRΔT/2M
ΔU= 1,5*2*8,31*100/3= 831Дж

Answer 2

Calculation of the Increase in Internal Energy of an Ideal Monoatomic Gas

To calculate the increase in internal energy of an ideal monoatomic gas when 2/3 of a mole of the gas is heated by 100 K, we can use the equation:

ΔU = (3/2) * n * R * ΔT

Where: - ΔU is the change in internal energy - n is the number of moles of the gas - R is the gas constant - ΔT is the change in temperature

In this case, we are given that 2/3 of a mole of the gas is heated by 100 K. Let's substitute the values into the equation:

ΔU = (3/2) * (2/3) * R * 100

To find the value of R, we can refer to the gas constant, which is approximately 8.314 J/(mol·K).

ΔU = (3/2) * (2/3) * 8.314 * 100

Simplifying the equation:

ΔU = 2 * 8.314 * 100

ΔU = 1662.8 J

Therefore, the increase in internal energy of the ideal monoatomic gas, when 2/3 of a mole of the gas is heated by 100 K, is approximately 1662.8 J.

Please note that the calculation assumes an ideal monoatomic gas and uses the ideal gas law.