Question

gas cylinder is able to withstand a maximum pressure of 1.20 x 10^6 Pa the cylinder has a volume of 3.00 X 10^-4 m3 it contains 3.00 x 10^22 molecules of gas. calculate using the ideal gay equation the temperature of the gas at maximum pressure

Answer 1

$$pV=\nu RT=\frac{N}{N_A}RT$$

$$(1.2\cdot 10^{6})(3\cdot 10^{-4})=\frac{3\cdot 10^{22}}{6.02\cdot 10^{23}}(8.31)T$$

$$T=869\ K$$