Algebra Problem!

Question

0.3 x 10⁴ liters of a ideal gas under a pressure of 0.0006 x 10⁶ atmospheres had a temperature of 3500 x 10⁷ K. After the volume was doubled and the pressure increased to 0.03 x 10⁵ atmospheres, what was the temperature of this ideal gas?

(A) 3.5x10⁹

(B) 3.5x10⁸

(C) 3.5x10¹¹

(D) 3.5x10¹²

(E) None of these

Answer 1

P₁V₁ / T₁ = P₂V₂ / T₂

(6 x 10²)(3 x 10³) / (3.5 x 10¹⁰) = (3 x 10³)(6 x 10³) / T₂

18 x 10⁶ / 3.5 x 10¹⁰ = 18 x 10⁶ / T₂

5.14 x 10^-4 = 1.8 x 10⁷ / T₂

5.14 x 10^-4 T₂ = 1.8 x 10⁷

T₂ = 1.8 x 10⁷ / 5.14 x 10^-4

T₂ = 0.35 x 10¹¹

Answer 2

P1 V1 = nR* T1

P1*V1

--------- nR <--- the # of moles and the gas constant are the same

T1

P1*V1 P2*V2

--------- = ----------- <--- n = # of moles remains the same; gas constant cancels out

T1 T2

plugs in the values; pressure is double and volume increased

(0.0006 x 10^{6) (}0.3 x 10^{4 ) (}0.03 x 10⁵⁾⁽0.6 x 10^{4 )}

----------------------------------- = ---------------------------------

3500 x 10⁷ T

Solves for T by cross multiplication:

T * (0.0006 x 10^{6) (}0.3 x 10^{4 ) = (}3500 x 10⁷⁾⁽0.03 x 10⁵⁾⁽0.6 x 10^{4 )}

^{T = (}3500 x 10⁷⁾⁽0.03 x 10⁵⁾⁽0.6 x 10^{4 )/}(0.0006 x 10^{6) (}0.3 x 10^{4 )}

^{= 350000 x 10 ^(7+5+4-6-4)}

= 350000 x 10^(2)

= 35000000

= 3.5 x 10^7

None of them