Question

Find the temperature in °C at which volume and pressure of 1 mol of nitrogen gas becomes 10 dm³ and 2.46 atmosphere respectively.

Answer 1

P = 2.46 atm, V = 10 dm³, n = 1 mol, R = 0.0821 dm³-atm K^-1 mol^-1
To find: Temperature (T)
Formula: PV = nRT
According to ideal gas equation,
PV = nRT
∴ T = \(\frac{\mathrm{PV}}{\mathrm{nR}}=\frac{2.46 \times 10}{1 \times 0.0821}\)
T = 299.63 K
Temp, in °C = 299.63 – 273.15 = 26.48 °C
Ans: The temperature of the nitrogen gas under the given conditions is 26.48 °C.