What is the total kinetic energy of translation in ergs of two moles of a perfect gas at 27 degree Celsius in calories?

To calculate the total kinetic energy of translation for two moles of a perfect gas at 27 degrees Celsius, we can use the formula ( KE = \frac{3}{2} nRT ), where ( n ) is the number of moles, ( R ) is the gas constant (approximately 1.987 cal/(mol·K)), and ( T ) is the temperature in Kelvin (300 K for 27°C). Thus, ( KE = \frac{3}{2} \times 2 , \text{mol} \times 1.987 , \text{cal/(mol·K)} \times 300 , \text{K} \approx 597.3 , \text{cal} ). To convert this to ergs, use the conversion ( 1 , \text{cal} = 4.184 \times 10^7 , \text{ergs} ), yielding approximately ( 2.50 \times 10^{10} , \text{ergs} ).