ceciliaxu27 Posted March 9, 2017 Report Share Posted March 9, 2017 what is the pressure , in pa, in a 100cm3 container containing 1.8g of steam at a temperature of 727 C?(R=8.31 JKmol) The key is (1.8)(8.31)(1000)/(18 * 10^-4) but I don'e understand that where is temperature..the formula is PV=nRT right? Also what is 1000 standard for and also 10 to the power of -4? Reply Link to post Share on other sites More sharing options...
Grace Taylor Posted March 9, 2017 Report Share Posted March 9, 2017 T= 1000 k (727C+273) and I believe volume should be m^3 Reply Link to post Share on other sites More sharing options...
kw0573 Posted March 10, 2017 Report Share Posted March 10, 2017 PV = nRT ==> P = nRT/V, n is in moles, but to find n we do mass/molar mass = 1.8 g divided by water's molar mass (18 g/mol) R = 8.31 J/mol/K T is in Kelvin, which is T(K) = T(°C) + 273 = 727 + 273 = 1000 K Finally, 1 m³ = 100 cm * 100 cm * 100 cm = 1 000 000 cm³ hence 100 cm³ = 1 * 10-4m³. We make conversion because 1 Pa = 1J / m³ Plug everything into P = nRT / V nRT / v = (1.8g * 8.31 J/mol/K * 1000 K) / (18 g/mol * 1 * 10-4m³) = (1.8)(8.31)(1000)/(18 * 10-4) J / m³ so numbers and units all match. Reply Link to post Share on other sites More sharing options...
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