Question

how many moles of gas is in the above 20.00 l container contains a gas at 50.00 °c and 1.000 atm. this is a missing variable problem, we need to define the variables: v = 1 l t = 2 k p = 3 atm n = ? r = 0.08205 l atm/mol k solving for n n = \frac{p}{r} putting in the numbers n = \frac{6 atm 7 l}{9 10 k} = 8 mol a. n₂ b. 20.0 c. 473.15 d. 323.15 e. 1.000 f. p g. 0 h. v i. 0 j. n₁ k. t l. 0 m. 1.464 n. 0.08205 l atm/mol k o. 0.7543 p. 0.3270 q. 12.6 r. 0.333 s. 3.000 t. 0.667 u. 99.38 v. 298.15 w. 2.000 x. 6.000 y. 4.000 z. 1.333 aa. 0.3270 bb. 7.329 cc. 22.414 dd. 32.00 ee. 10.46 ff. 1.428 gg. r hh. 273.15

Explanation:

Step1: Convert temperature to Kelvin

To convert Celsius to Kelvin, use the formula \(T(K)=T(^{\circ}C)+273.15\). Given \(T = 50.00^{\circ}C\), then \(T=50.00 + 273.15=323.15\) K.

Step2: Identify variables

From the problem, \(V = 20.00\) L (so \(1\) is \(B\)), \(T = 323.15\) K (so \(2\) is \(D\)), \(P = 1.000\) atm (so \(3\) is \(E\)).

Step3: Use ideal gas law formula

The ideal gas law is \(PV=nRT\), solving for \(n\) gives \(n=\frac{PV}{RT}\). Here, the formula for \(n\) has \(P\) in the numerator (so \(4\) is \(H\)) and \(T\) in the denominator (so \(5\) is \(K\)).

Step4: Substitute values

Substitute \(P = 1.000\) atm (so \(6\) is \(E\)), \(V = 20.00\) L (so \(7\) is \(B\)), \(R=0.08205\) L atm/mol K (so \(9\) is \(N\)), \(T = 323.15\) K (so \(10\) is \(D\)) into \(n=\frac{PV}{RT}\). Then \(n=\frac{1.000\times20.00}{0.08205\times323.15}\).

Step5: Calculate \(n\)

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Answer:

1. B. 20.0
2. D. 323.15
3. E. 1.000
4. H. V
5. K. T
6. E. 1.000
7. B. 20.0
8. O. 0.7543
9. N. 0.08205 L atm/mol K
10. D. 323.15