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If formals of gas are added to a container that already hold one mole of gas, how will the pressure change with in the container?
Assuming "Four Moles" was meant, the total new amount is 5 moles. From the , at constant temperature: $5/1 = P_2/P_1$ So the new pressure will be five times the...
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If 4 moles of gas are added to a container that already holds 1 mole of gas, what happens to the pressure inside the container due to the change in the amount of gas?
At constant volume, and constant temperature, pressure is proportional to the number of molecules, and thus to the number of moles. We can see this immediately from the Ideal...
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Container A holds 757 mL of ideal gas at 2.60 atm. Container B holds 164 mL of ideal gas at 4.80 atm. If the gases are allowed to mix together, what is the resulting pressure?
The ideal gas equation of state is $PV = nRT$ In this problem we may consider the temperature to be constant, so we can set up an equation that...
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A 9.70 L container holds a mixture of two gases at 53C. The partial pressures of gas A and gas B, respectively, are 0.368 atm and 0.893 atm. If 0.210 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
$"Dalton's Law of Partial Pressures......"$, $"which states, that in a"$ $"gaseous mixture, the partial pressure of a gaseous component is"$ $"the same as the pressure it would exert if it...
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A 7.35-L container holds a mixture of two gases at 17 °C. The partial pressures of gas A and gas B, respectively, are .415 atm and .720 atm. If .160 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
Since pressures are additive, and we know $P_A$ and $P_B$, all we need to do is calculate $P_C$. $P_C = (n_CRT)/V$ $=$ $(0.160cancel(*mol)xx0.0821cancel(*L)(*atm)*cancel(K^-1)*cancel(mol^-1)xx300cancel(K))/(7.35cancel(L))$ $=$ $??$ $atm$ $P_(TOTAL) = P_C +...
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Container A holds 737 mL of ideal gas at 2.30 atm. Container B holds 114 mL of ideal gas at 4.40 atm. lf the gases are allowed to mix together, what is the resulting pressure?
According to Dalton's Law of Partial Pressures, each gas will exert its pressure independently of the other. Hence we can use to calculate the pressure of each gas separately...
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A container holds 6.4 moles of gas. Hydrogen gas makes up 25% of the total moles in the container if the total pressure is 1.24 atm. What is the partial pressure of hydrogen?
$P_"Total"=(n_"Total"RT)/V$ $=(n_(H_2)+n_"other gases")/V$ But $n_(H_2)=25%xxn_"Total"$. And thus $P_(H_2)=25%xxP_"Total"=25%xx1.24*atm=??$ For more spray, see This treatment relies on $"Dalton's Law of Partial Presssures"$, which states that in $"in a gaseous...
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What is the partial pressure of $"CO"_2$ in a container that holds 5 moles of $"CO"_2$, 3 moles of $"N"_2$, and 1 mole $"H"_2$ and has a total pressure of 1.05 atm?
$p("CO"_2)="Total pressure"*"Mole fraction of CO"_2=1.05*5/(5+3+1)=1.05*5/9=0.58"atm"$
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A flexible container holds $"6.70 mol"$ of a gas. An additional $"3.50 mol"$ of the gas is added to the container until it reaches a final volume of $"25.4 L"$ while temperature and pressure are kept constant, what was the initial volume?
This problem requires , which states that the amount (moles) of a gas is directly proportional to its volume, such that if the amount increases, the volume increases and vice...
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2 moles of H2O(g), 2.6 moles of N2(g), 2.00 moles of H2(g), and 2.00 moles of NO(g) are in equilibrium in a 2 L container. If a second 1L container with 1 mole of H2(g) is combined with the first, what is the initial reaction quotient?
We know that the first container is $2$ liters in volume, and the one filled with hydrogen is $1$ liter, so the combined volume is $color(green)(3$ $color(green)("L"$. And after...
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