Question

A sample of carbon monoxide gas occupies a volume of 263 mL at a pressure of 781.3 torr and a temperature of 399 K. What woul

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Answer #1

We have, P V = n R T

where, P is a pressure of a gas, V is a volume of a gas, n is no. of moles of gas, R is a gas constant and T is temperature of gas.

Given

P 1 V 1 = n 1 R T 1   \rightarrow    P 2 V 2 = n 2 R T 2

P 1 = 781.3 torr

V 1 = 263 ml

T 1 = 399 K

P 2 = 657.5 torr

V 2 = 82 ml

T 2 = ?

In above case, no. of moles gas are constant i e n 1 = n 2

We can write, P 1 V 1 / P 2 V 2 = n 1 R T 1 / n 2 R T 2

\therefore  P 1 V 1 / P 2 V 2 =  T 1 / T 2

( 781.3 torr \times 263 ml ) / ( 657.2 torr \times 82 ml ) = 399 K / T 2

3.81 = 399 K / T 2

T 2 = 399 K / 3.81 = 104.7 K

ANSWER : 104.7 K

PART 2

We have,  P 2 V 2 / P 1 V 1 = n 2 R T 2 / n 1 R T 1

At constant temperature and volume, we get P 2 /P 1   =   n 2 / n 1

Therefore, 2.3 atm / 1.00 atm =  n 2 / 2.7 mol

2.3 = n 2 / 2.7 mol

n 2 = 2.3 ( 2.7 mol) = 6.21 mol

No. of Moles of gas added = Final moles of gas - Initial moles of gas = 6.21 - 2.7 = 3.51 mol

ANSWER : No. of Moles of gas added = 3.51 mol

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