Dare: [ mu=29 cdot 10^-3 mathrm(kn) p=10^5 mathrm(Tla) V=3.5 cdot 10=150 mathrm(~m)^3 t=23^circ mathrm(C) m-? ]

To find the mass (m), we can use the ideal gas law equation:<br /><br />\[ PV = \mu RT \]<br /><br />Where:<br />- P is the pressure (in Pa)<br />- V is the volume (in m³)<br />- μ is the absolute pressure (in Pa)<br />- R is the specific gas constant (for air, R ≈ 287 J/(kg·K))<br />- T is the temperature (in K)<br /><br />First, let's convert the given values to appropriate units:<br />- μ = 29 × 10⁻³ kN = 29 × 10³ N = 29,000 N<br />- P = 10⁵ Tla = 10⁵ × 10¹² Pa = 10¹⁷ Pa<br />- V = 3.5 × 10 = 150 m³<br />- T = 23°C = 23 + 273.15 K = 296.15 K<br /><br />Now, we can rearrange the ideal gas law equation to solve for m:<br /><br />\[ m = \frac{PV}{\mu RT} \]<br /><br />Substituting the given values:<br /><br />\[ m = \frac{(10^{17} \, \text{Pa}) \times (150 \, \text{m}^3)}{(29,000 \, \text{N}) \times (287 \, \text{J/(kg·K)}) \times (296.15 \, \text{K})} \]<br /><br />\[ m \approx 1.94 \times 10^3 \, \text{kg} \]<br /><br />Therefore, the mass (m) is approximately 1,940 kg.