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Lera25 [3.4K]
3 years ago
9

The ideal gas constant, R has several different values that could be used. Which quantity causes these differences? pressure tem

perature volume moles
Chemistry
1 answer:
Blababa [14]3 years ago
3 0

Answer : The correct option is, pressure.

Explanation :

The ideal gas equation is,

PV=nRT

where,

P = pressure of the gas

V = volume of the gas

n = number of moles of gas

T = temperature of the gas

R = gas constant

The value of 'R' has several different values which are :

R=0.08205L.atm/mol.K

R=8.3145L.kPa/mol.K

R=8.3145J/mol.K 

R=1.987cal/mol.K

R=0.08314L.bar/mol.K

That means, the value of 'R' is different due the change in the pressure value and all the variables (temperature, volume and moles) are constant.

Hence, the correct option is, pressure.

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A helium-filled balloon at 310.0 K and 1 atm, contains 0.05 g He, and has a volume of 1.21 L. It is placed in a freezer (T = 235
trapecia [35]

Answer : The value of \Delta E of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}

Molar mass of helium = 4 g/mole

\text{Moles of helium}=\frac{0.05g}{4g/mole}=0.0125mole

Now we have to calculate the heat.

Formula used :

q=nc_p\Delta T\\\\q=nc_p(T_2-T_1)

where,

q = heat

n = number of moles of helium gas = 0.0125 mole

c_p = specific heat of helium = 20.8 J/mol.K

T_1 = initial temperature = 310.0 K

T_2 = final temperature = 235.0 K

Now put all the given values in the above formula, we get:

q=nc_p(T_2-T_1)

q=(0.0125mole)\times (20.8J/mol.K)\times (235.0-310.0)K

q=-19.5J

Now we have top calculate the work done.

Formula used :

w=-p\Delta V\\\\w=-p(V_2-V_1)

where,

w = work done

p = pressure of the gas = 1 atm

V_1 = initial volume = 1.21 L

V_2 = final volume = 0.99 L

Now put all the given values in the above formula, we get:

w=-p(V_2-V_1)

w=-(1atm)\times (0.99-1.21)L

w=0.22L.artm=0.22\times 101.3J=22.29J

conversion used : (1 L.atm = 101.3 J)

Now we  have to calculate the value of \Delta E of the gas.

\Delta E=q+w

\Delta E=(-19.5J)+22.29J

\Delta E=2.79J

Therefore, the value of \Delta E of the gas is 2.79 Joules.

3 0
3 years ago
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