How much heat does it take to increase the temperature of 2.70 mol of an ideal gas by 30.0 K near room temperature if the gas is held at constant volume and is diatomic

1683.6J

Explanation:

Given:

n= no. Of mol= 2.70 mol

T= Temperature= 30.0 K

Q= n Cv × ∆T eqn(1)

Where CV= molar heat capacity=5/2R for diatomic particle ,such as H2

CV= molar heat capacity=3/2R for diatomic, such as H

R= gas constant= 8.314 J/mol.K

Q= heat energy

For a diatomic molecules

Q= n Cv × T

But

Cv= molar heat capacity=5/2R = 5/2(8.314)=20.785

CV= 20.785

. ∆T= Temperature= 30.0 K

Then substitute the values into the eqn(1)

Q= 2.70 × 5/2(8.314) × 30

Q= 2.70 × 20.785 × 30

=1683.6J

answer; i believe the correct answer is (c) /// is a homogeneous mixture of gaseous substance nitrogen,oxygen and smaller amounts of other substancesmixtures can be either ( homogeneous or heterogeneous)plz check first good luck

i'm pretty sure its b i might be wrong though!