One mole of an ideal monoatomic gas at temperature T0 expands slowly according to the law (P/V). If the final temperature is 2T0 heat supplied the gas is

Question

One mole of an ideal monoatomic gas at temperature T0 expands slowly according to the law (P/V). If the final temperature is 2T0 heat supplied the gas is (A) 2RT0 (B) RT0 (C) (3/2)RT0 (D) (1/2)T0

Tardigrade Admin · Accepted Answer

In a process PVx = constant, molar heat capacity is given by C=(R/&gamma;-1)+(R/1-x) As the process is (P/V)= constant, i.e., PV-1 = constant, therefore, x =-1. For an ideal monoatomic gas, &gamma;=(5/3) ∴ C=(R/(5)3-1)+(R/1-(-1))=(3/2)R+(R/2)=2 R ÎQ = nC(ÎT) = 1(2R)(2T0 - T0) = 2RT0

Q. One mole of an ideal monoatomic gas at temperature $T_{0}$ expands slowly according to the law $\frac{P}{V} .$ If the final temperature is $2 T_{0}$ heat supplied the gas is

$2 R T_{0}$

$R T_{0}$

$\frac{3}{2} R T_{0}$

$\frac{1}{2} T_{0}$

Q. One mole of an ideal monoatomic gas at temperature T0​ expands slowly according to the law VP​. If the final temperature is 2T0​ heat supplied the gas is

2RT0​

RT0​

23​RT0​

21​T0​

In a process PVx= constant, molar heat capacity is given by C=γ−1R​+1−xR​ As the process is VP​= constant, i.e., PV−1= constant, therefore, x=−1. For an ideal monoatomic gas, γ=35​ ∴C=35​−1R​+1−(−1)R​=23​R+2R​=2R ΔQ=nC(ΔT)=1(2R)(2T0​−T0​)=2RT0​

Q. One mole of an ideal monoatomic gas at temperature $T_{0}$ expands slowly according to the law $\frac{P}{V} .$ If the final temperature is $2 T_{0}$ heat supplied the gas is

$2 R T_{0}$

$R T_{0}$

$\frac{3}{2} R T_{0}$

$\frac{1}{2} T_{0}$