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dolphi86 [110]
3 years ago
9

A sample of Ar gas has a volume of 6.30 L with an unknown pressure. The gas has a volume of 8.57 L when the pressure is 2.74 atm

, with no change in temperature or amount of gas. What was the initial pressure, in atmospheres, of the gas?
Chemistry
1 answer:
ExtremeBDS [4]3 years ago
8 0

Answer:

3.73 atm

Explanation:

Since there is no change in temperature, we use Boyle's law.

P₁V₁ = P₂V₂ where P₁ = initial pressure of Ar gas = unknown, P₂ = final pressure of Ar gas = 2.74 atm, V₁ = initial volume of Ar gas = 6.30 L and V₂ = final volume of Ar gas = 8.57 L

P₁ = P₂V₂/V₁ = 2.74 atm × 8.57 L/6.30 L = 3.73 atm

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One cubic meter of an ideal gas at 600 K and 1000 Malek kPa expands to five times its initial volume as follows; a) By a mechani
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Answer:

Follows are the solution to this question:

Explanation:

Given value:

v_1=1m^3 \\ t_1 = 600 \ K \\ p_1 = 1000 \ kpa \\v_1 = 5 \ v_1  = 5 \ m^3

In point a:

Calculating the process of Isothermal, when the temperature is constant:

\to  T_1 = T_2 = 600 \ K \\\\\to \bold{P_1V_1 = P_2V_2} \\\\\to 1000 \ (Kpa)  \times 1 \ (m^3) \neq p_2 \times  5 \ (m^3)\\\\\to p_2 = 200 \ kpa\\\\\to w = nRT \ In (\frac{v_2}{v_1}) =  p_1v_1   \ ln ( \frac{v_2}{v_1})

                               = 1000 \times  1 \times ln  (\frac{5}{1}) \\\\ = 1.61 \ KJ

In point b:

Calculating the adiobatic process:

\to  p_1v_1^\gamma  = p_2v_2^\gamma \\\\ \to \gamma  = \frac{c_p}{c_v} \\\\\to  R= c_p -c_v  \\\\ \to c_p= 21 \frac{J}{mol.k}\\\\\to \gamma  = \frac{c_p}{c_p-R} \\\\

       = \frac{21}{21.8}\\\\ = 1.62\\\\= 1000 \times 1^{1.62}\\\\ = p_2 \times  5^{1.62}\\\\

p_2 =  73.73 \ Kpa

\to p_1^{1-\gamma} t_1^{\gamma} = p_2^{1-\gamma } t_2^{\gamma }

= 1000^{1-1.62} \times 600^{1.62} = 73.73^{1-1.62} \times t_2^{\gamma}\\\\ \to t_2^{\gamma} = 6288.5\\\\\to t_2= 6258.5 ^\frac{1}{1.62} \\\\

        = 221.2 \ k

In point c:

\to w= \frac{p_2v_2-p_1v_1}{\gamma -1}

        = \frac{(73.73 \times 5)- ( 1000\times 1)}{1.620-1}

        = \frac{( -631.35 )}{.620}\\\\= -1018.31 \ KJ

workdone by gas is 1.018 KJ

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