To find the total energy of one mole of an ideal monatomic gas, we need to use the formula related to the internal energy of the gas. For an ideal monatomic gas, the internal energy ($U$) is given by the expression:

$$ U = \frac{3}{2} nRT $$

where:

$n$ is the number of moles of the gas,
$R$ is the ideal gas constant, and
$T$ is the temperature in Kelvin.

Given that $n = 1$ mole and the temperature is $27^\circ C$, we first need to convert this temperature to Kelvin. The conversion from Celsius to Kelvin is done by adding 273:

$$ T = 27 + 273 = 300 \, \text{K} $$

The ideal gas constant $R$ can be used in various units, but since we want the energy in calories, we will use the value of $R = 2 \, \text{cal/mol}\cdot\text{K}$. Plugging these values into the formula, we get:

$$ U = \frac{3}{2} \times 1 \times 2 \times 300 \, \text{cal} $$

$$ U = \frac{3}{2} \times 600 \, \text{cal} $$

$$ U = 900 \, \text{cal} $$

Thus, rounding to an appropriate number of significant figures, the total energy of one mole of an ideal monatomic gas at 27°C is approximately 900 calories.

JEE Advanced Subjects