(b) PV =RT The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will be 5 (a) PV = {RT (b) PV RT 16 2 5 (c) PV = RT 32 (d) PV = 5 RT -
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- The equation of state for g of oxygen at a pressure P, and temperature T, when occupying a volume V, will be? a) PV = (5/32)RT b) PV = 5RT c) PV = (5/2)RT d) PV = (5/16)RTProblem 3. The viral coefficients of a gas at 20 °C and 11.5 bar are B = -138 cm³ mol¹ and C=7222 cmº mol². Calculate the V (molar volume) Z (compressibility factor) of the gas. Use the equation below (R = 83.14 cm³ bar mol-¹ K-¹). PV 2 = ² = (1 + = + =) Z RTAn aluminum coffee mug has in inner diameter of 2.50 cm at room temperature (20.0° C). If someone accidentally leaves the cup on the stove burner its temperature could rise to 300° C. What would be the inner diameter then? (keep 4 significant figures in your answer) Linear coefficient of expansion for copper, Caluminum = 24 X 10-6/C° Area coefficient of expansion for copper, Yaluminum = 48 X 10-/C° Volume coefficient of expansion for copper, Baluminum = 72 X 106/C°
- The figure (not to scale) shows a pV diagram for 10 dg of helium gas (He) that undergoes the process 1 → 2 → 3. Find the value of V3 and V1. The ideal gas constant is R = 8.314 J/mol∙K = 0.0821 L∙atm/mol∙K, and the atomic weight of helium is 4.0 g/mol.2 mol of an ideal monoatomic gas moves from State 1 to State 2 P at constant pressure 1000 Pa and size V1=2 m3,V2 =3 m3. Calculated value W, Q, AU, TI, T2 P Solution: Va V2 P A.For any gas, C C. - (+7), (7), Suppose you have one mole of a gas that obeys the equation of state, p(v- b) = represents the volume taken up by the molecules in the system, and 3.5 bars and T = 425 K for your gas sample. Hint: Use the equation of state to evaluate the partial derivatives. =RT, where b is the molar volume. Find the value of A if C-C₁=AR when P = 6
- (1) For a gas that follows equation of state pV = f(T), show that дру 1 df V dT' (ӘТ V (N)₂ P ƏT = 1 df p dTFind the K.E. (1) per cm³ (2) per mole (3) per gram and (4) per molecule of nitrogen at N.T.P. (Molecular weight of N₂ 28, Normal pressure = 76 cm of mercury, density of Hg 13.6 gm/cm³ g = 980 cm/s², Avogadro's number 6.023 x 1023 molecules/mole. (R = 8.314 x 10² erg/mole K)The vapor pressures of the components, A and B, in a binary solution have been modeled and found to obey хара exp(0.75 XB) A A exp(0.75x) where XÃ and are the mole fractions, and PA* and PB* are the vapor pressures of each pure substance at room temperature. (a) If PA* = 0.084 bar and the total pressure of a mixture with XA = 0.40 is 0.125 bars, what is PB*, the vapor pressure of pure B (in bars)? P = X P А P₁ = X_P B QUESTION 14 B B * * Continuation of the previous problem (b) Assuming that the vapor is an ideal gas, what is the mole fraction of component B in the vapor phase?
- α = alpha A possible equation of state for a gas takes the formPV = RT * exp (-α / VRT)in which α and R are constants. Calculate expressions for(∂P/∂V)T, (∂V/∂T)P, (∂T/∂P)V,and show that their product is −1.Derive an explicit equation for the change in internal energy of a van del Waals gas with the equation of state in the form: P =RT/(v − b) − a/v^2Assume cv varies linearly with temperature, cv = c1 + c2TCalculate the internal energy, U internal -in Joules- of 3.2 moles of a monoatomic gas, if 3 Cv, monoatomic = R The temperature of the gas is T = 349K R = 8.31 J mol. K Hint: U₁ internal = n. Cy. T