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Question: 1/2 mole of helium is contained in a container at STP. How much heat energy is needed to double the ...

1/2 mole of helium is contained in a container at STP. How much heat energy is needed to double the pressure of the gas (volume is constant) heat capacity of has is 3J/g/K
(A) 1436
(B) 736
(C) 1638
(D) 5698

Explanation

Solution

The pressure of a given mass of gas is directly proportional to the absolute temperature provided that the volume is kept constant. Use that relation to calculate the temperature difference of the two different states of the system (the high pressure state and the stp pressure state). Standard temperature is considered as zero degree Celsius or 273 Kelvin.
Formula used: In this solution we will be using the following formulae;
P=kTP = kT where PP is the pressure of an ideal gas at a certain state, TT and is the absolute temperature of the gas at the same state, kk is a proportionality constant.
Q=nMcvΔTQ = nM{c_v}\Delta T where QQ is the heat absorbed by a gas,MM is the molar mass,nn is the number of moles of the gas, cv{c_v} is the specific heat capacity at constant volume and ΔT\Delta T is the difference in temperature after a particular amount of absorbed heat.

Complete Step-by-Step solution:
To calculate for heat needed, we must first calculate the temperature necessary for the condition to hold.
Generally, it is known that the pressure of a gas is directly proportional to temperature of the gas provided the volume is held constant. Hence
P=kTP = kT where PP is the pressure of an ideal gas at a certain state, TT and is the absolute temperature of the gas at the same state, kk is a proportionality constant.
Hence, for the first state, we have
P1=kT1=273k{P_1} = k{T_1} = 273k (since the standard temperature is 373 K)
k=P1273\Rightarrow k = \dfrac{{{P_1}}}{{273}}
For the second state, we have
2P1=kT2\Rightarrow 2{P_1} = k{T_2}
Inserting the known kk expression, we have
2P1=(P1273)T22{P_1} = \left( {\dfrac{{{P_1}}}{{273}}} \right){T_2}
2=T2273\Rightarrow 2 = \dfrac{{{T_2}}}{{273}}
Hence, we have
T2=2×273=546K{T_2} = 2 \times 273 = 546K
Now, the heat required can be given by
Q=nMcvΔTQ = nM{c_v}\Delta T where nn is the number of moles of the gas,MM is the molar mass,cv{c_v} is the specific heat capacity at constant volume and ΔT\Delta T is the difference in temperature after a particular amount of absorbed heat.
Hence, we have
Q=(0.5)4(3)(546273)Q = \left( {0.5} \right)4\left( 3 \right)\left( {546 - 273} \right)
By computation,
Q=1638JQ = 1638J

The correct option is C

Note: Alternatively, instead for calculating for kk and then using its value in the second state, we could simply divide state 1 by state 2, and hence, we have
P1P2=T1T2\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{T_1}}}{{{T_2}}}
Then by making T2{T_2} subject of formula, we have
T2=P2T1P1{T_2} = \dfrac{{{P_2}{T_1}}}{{{P_1}}}
Hence, by inserting the known values and expressions, we have
T2=2P1(273)P1=546K{T_2} = \dfrac{{2{P_1}\left( {273} \right)}}{{{P_1}}} = 546K
Which is identical to what is gotten above.