Question

The difference between the principal specific heats of nitrogen is 300 J/kg °K and ratio of the two specific heats is 1.4. then the CP is
1050 J/kg °K
650 J/kg °K
750 J/kg °K
150 J/kg °K
17
CP
and
CV
denote the molar specific heats of a gas at constant pressure and at constant volume respectively. If and
CP/CV=γ,
then
CV
is equal to

R/(γ−1)

(γ−1)/R

Nosolution

373.2oC
18 A gas is taken in a sealed container at 300 K. it is heated at constant volume to a temperature 600 K. the mean K.E. of its molecules is
Halved
Doubled
Tripled
Quadrupled
19 R.M.S velocity of a gas molecule of mass m at given temperature is proportional to

mo
m

m1/2

1/m1/2
20 The mean kinetic energy of one gram-mole of a perfect gas at absolute temperature T is
0.5kT
0.5RT
1.5kT
1.5RT

Answer 1

Answer on Question #51926-Physics-Field Theory

The difference between the principal specific heats of nitrogen is $300\mathrm{J / kg}{}^{\circ}\mathrm{K}$ and ratio of the two specific heats is 1.4. Then the CP is

1050 J/kg °K

650 J/kg °K

750 J/kg °K

150 J/kg °K

Solution

$$C _ {P} - C _ {V} = 3 0 0 \frac {\mathrm {J}}{\mathrm {k g} {}^ {\circ} \mathrm {K}}; \frac {C _ {P}}{C _ {V}} = 1. 4.$$ $$C _ {P} \left(1 - \frac {1}{\frac {C _ {P}}{C _ {V}}}\right) = C _ {P} - C _ {V}.$$ $$C _ {P} = \frac {C _ {P} - C _ {V}}{\left(1 - \frac {1}{\frac {C _ {P}}{C _ {V}}}\right)} = \frac {3 0 0 \frac {\mathrm {J}}{\mathrm {k g} {}^ {\circ} \mathrm {K}}}{1 - \frac {1}{1 . 4}} = 1 0 5 0 \frac {\mathrm {J}}{\mathrm {k g} {}^ {\circ} \mathrm {K}}.$$

Answer: 1050 J/(kg°K).

17 CP and CV denote the molar specific heats of a gas at constant pressure and at constant volume respectively. If and CP/CV=γ, then CV is equal to

R/(γ-1)

(γ-1)/R

Nosolution

373.2oC

Solution

$$C _ {P} - C _ {V} = R; \frac {C _ {P}}{C _ {V}} = \gamma .$$ $$C _ {V} \left(\frac {C _ {P}}{C _ {V}} - 1\right) = R.$$ $$C _ {V} = \frac {R}{(\gamma - 1)}.$$

Answer: R/((γ-1)).

18 A gas is taken in a sealed container at $300\mathrm{K}$. It is heated at constant volume to a temperature $600\mathrm{K}$. The mean K.E. of its molecules is

Halved

Doubled

Tripled

Quadrupled

**Solution**

The mean K.E. of gas molecules $\sim T$. Thus $\frac{T'}{T} = \frac{600K}{300K} = 2$. So, it is doubled.

**Answer: Doubled.**

19 R.M.S velocity of a gas molecule of mass m at given temperature is proportional to

mo

m

$$m^{\left(\frac{1}{2}\right)}$$ $$\frac{1}{m^{\left(\frac{1}{2}\right)}}$$

**Solution**

R.M.S velocity of a gas molecule of mass m is

$$v_{rms} = \sqrt{\frac{3kT}{m}}.$$

So, R.M.S velocity of a gas molecule of mass m at given temperature is proportional to $\frac{1}{m^{\left(\frac{1}{2}\right)}}$.

**Answer: $1/m^{\prime}(1/2)$**

20 The mean kinetic energy of one gram-mole of a perfect gas at absolute temperature T is

0.5kT

0.5RT

1.5kT

1.5RT

**Solution**

$$KE_{avg}(per mole) = \frac{3}{2} RT.$$

So, the mean kinetic energy of one gram-mole of a perfect gas at absolute temperature $T$ is 1.5RT.

Answer: 1.5RT.

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