Raj and Tonya each participated in an experiment

Question

Raj and Tonya each participated in an experiment designed to demonstrate the ideal gas law, PV = nRT. In this equation , P is pressure, V is volume of the container, n is a measure of the amount of gas, R is a constant, and T is the temperature. Tonya filled her 1-liter container completely with argon gas that the lab instructor provided at a temperature of 300° Kelvin and measured the pressure as 140 kilopascals. She then transferred the argon through a hose to Raj's 2-liter container, but they suspected that some gas had leaked out in the process. Raj measured the pressure in his container as 60 kilopascals and the temperature as 280° Kelvin. What fraction of the argon was lost in the transfer from Tonya to Raj?

A) 4/105R
B) 4/49R
C) 4/49
D) 1/7
E) 7/15

A) 4/105R
B) 4/49R
C) 4/49
D) 1/7
E) 7/15

nick1816 · Accepted Answer

\frac{PV}{nT } = constant

\frac{P_1V_1}{n_1T_1}= \frac{P_2V_2}{n_2T_2}

\frac{140*1}{n_1*300} = \frac{60*2}{n_2*280}

\frac{n_2}{n_1} = \frac{45}{49}

Fraction of gas leaked = 1-(45/49) = 4/49

jabhatta2 · Answer

Bunuel   VeritasKarishma   chetan2u  
 nick1816

Why are we doing N2 / N1

I did this by getting the ratio of N1/ N2

Please suggest how to determine whether you have to do N2/N1 or N1/N2

Thank you !

CrackverbalGMAT · Answer

For Tonya:  P1 = 140, V1 = 1, T1= 300, n1 = moles transferred from Tonya (we really don't need to know the unit of n1 and can be thought as units)

For Raj:  P2 = 60, V2 = 2, T2= 280, n2 = moles received by Raj

Since  \frac{PV}{nRT}  is a constant, hence  \frac{P_1 \space * \space V_1}{n_1 \space * \space T_1} = \frac{P_2 \space * \space V_2}{n_2 \space * \space T_2}  (R is a constant, it cancels out and so do Options 1 and 2)

Substituting the values, we get,  \frac{140 \space * \space 1}{n_1 \space * \space 300} = \frac{60 \space * \space 2}{n_2 \space * \space 280}

\frac{n_1}{n_2} = \frac{140 \space * \space 1 * \space 280}{60 \space * \space 2 * \space 300} = \frac{49}{45}

This indicates that when Tonya transferred 49 units, Raj received 45 units. So 49 - 45 = 4 units of the 49 transferred was lost =  \frac{4}{49}

Option C

Arun Kumar

KarishmaB · Answer

Question: What fraction of the argon was lost in the transfer from Tonya to Raj?

n1 is the amount Tonya had and n2 is the amount Raj had.

We get n1/n2 = 49/45 or n2/n1 = 45/49

This means that for every 49 parts that Tonya had, Raj got only 45 parts. So 4 parts were lost.
What fraction of the gas was lost? 4 parts out of 49 parts were lost. So 4/49 was lost.

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Raj and Tonya each participated in an experiment

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Raj and Tonya each participated in an experiment

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