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Question: For one mole of and ideal gas, the slope of \(V\)vs \(T\) curve at constant pressure of \(2atm\) is ...

For one mole of and ideal gas, the slope of VVvs TT curve at constant pressure of 2atm2atm is X Lmol1K1X{\text{ }}Lmo{l^{ - 1}}{K^{ - 1}}. The value of the ideal universal gas constant R'R' in terms of XX is:
A. X L atm mol1K1X{\text{ }}L{\text{ }}atm{\text{ }}mo{l^{ - 1}}{K^{ - 1}}
B. X2 L atm mol1K1\dfrac{X}{2}{\text{ }}L{\text{ }}atm{\text{ }}mo{l^{ - 1}}{K^{ - 1}}
C. 2X L atm mol1K12X{\text{ }}L{\text{ }}atm{\text{ }}mo{l^{ - 1}}{K^{ - 1}}
D. 2X atm L1 mol1K12X{\text{ }}atm{\text{ }}{L^{ - 1}}{\text{ }}mo{l^{ - 1}}{K^{ - 1}}

Explanation

Solution

The gas constant (universal gas constant) is denoted by symbol RR. It is equal to Boltzmann constant, but expressed in units of energy per temperature increment per mole i.e. the pressure- volume product, rather than energy per temperature, increment per particle.

Formula used: The ideal gas equation PV=nRT \Rightarrow PV = nRT

Complete step by step answer: The gas constant occurs in the ideal gas law as follows:
PV=nRT=mRspecificTPV = nRT = mRspecificT
Where, PP is absolute pressure (SI units pascals), VV is volume of the gas (SI unit cubic metre), nn is the amount of gas (SI unit moles), mm is mass (SI unit kilograms) contained in VV and TT is the thermodynamic temperature (SI unit Kelvins), RR specific is the mass specific gas constant. The gas constant is expressed in the same physical units as molar entropy and molar heat capacity.
RRspecific =RM=Cp=Cv = \dfrac{R}{M} = Cp = Cv
MM= molar mass of gas mixture
PV=nRT, or,P(vn)=RTPV = nRT,{\text{ }}or,P\left( {\dfrac{v}{n}} \right) = RT
Or PVm=RTPVm = RT
(Vm=molar volume)\left( {Vm = molar{\text{ }}volume} \right)
Vm=RPTVm = \dfrac{R}{P}T ………. (i)
At constant pressure for 11 mol of an ideal gas, V=RPTV = \dfrac{R}{P}T
The relation represents a straight line passing through the equation, so, for 11 mol of an ideal gas at constant pressure, the graph of VmVm vs TT, will be a straight line with slope R/PR/P.

It is given in the equation that RP=m=X\dfrac{R}{P} = m = X=given
RP=X\dfrac{R}{P} = X
R2atm=X L mol1K1\dfrac{R}{{2atm}} = X{\text{ }}L{\text{ }}mo{l^{ - 1}}{K^{ - 1}}
R=2X L atm mol1K1R = 2X{\text{ }}L{\text{ }}atm{\text{ }}mo{l^{ - 1}}{K^{ - 1}}.
So, the correct answer is “Option C”.

Note: The RR is known as gas constant or universal gas constant or ideal gas constant. It’s value is equal to approx. 8.31446261815324J K1mol18.31446261815324J{\text{ }}{K^{ - 1}}mo{l^{ - 1}}.